W9-W12. Infinite Series

Author

Mohammad Alkousa

Published

November 6, 2025

1. Theory

1.1 Sequences and Limits

A sequence is an ordered list of numbers where each is determined by some rule or formula. Understanding the limit of a sequence is fundamental: we say if the terms get arbitrarily close to the value as increases without bound.

1.1.1 Computing Limits Involving

Many limits involve the fundamental constant . More generally, we have the crucial identity:

When computing limits, we often need to manipulate expressions to match this form. Key techniques include:

Factorial simplification: Remember that
Power manipulation: Express ratios of powers as a single power, e.g.,
Rewriting to standard form: Transform expressions like into the form by appropriate substitutions
Exponent manipulation: Use the property to isolate the standard limit

1.1.2 The Squeeze Theorem

The Squeeze Theorem (or Sandwich Theorem) is a powerful tool for evaluating limits of sequences that oscillate or are difficult to compute directly. If we have three sequences satisfying for all sufficiently large , and if , then we can conclude that .

This is particularly useful for dealing with oscillating terms like , since we can bound them: .

1.1.3 Bounded and Monotonic Sequences

A sequence is:

Bounded above if there exists such that for all
Bounded below if there exists such that for all
Bounded if it is both bounded above and below
Monotonic increasing if for all
Monotonic decreasing if for all

The Bolzano-Weierstrass Theorem (also called the Monotone Convergence Theorem) states that every bounded monotonic sequence converges. This is extremely useful: if we can prove a sequence is bounded and monotonic, we immediately know it converges, even if finding the exact limit requires additional work.

1.1.4 Proving Properties by Induction

To prove that a recursively defined sequence satisfies certain properties (like being bounded or monotonic), we often use mathematical induction:

Base case: Verify the property holds for (or the first few terms)
Inductive hypothesis: Assume the property holds for
Inductive step: Prove that if the property holds for , it must also hold for

Once boundedness and monotonicity are established, we can find the limit by treating the recursive formula as an equation where both and approach the same limit .

1.2 Infinite Series

1.2.1 Basic Definitions

An infinite series (or simply series) is the sum of infinitely many terms from a sequence. We write this as:

While we cannot literally add infinitely many numbers, we can define what we mean by such a sum using the concept of partial sums. The -th partial sum is:

The series converges if the sequence of partial sums converges to a finite limit . In that case, we write and call the sum of the series. If the sequence of partial sums does not converge to a finite limit, we say the series diverges.

1.2.2 Telescoping Series

A telescoping series is one where most terms cancel out when we expand the partial sum. Formally, if we can write the general term as , then:

The series converges if and only if exists, and in that case:

A common technique is to use partial fraction decomposition to rewrite terms in a form that telescopes. For example, leads to a telescoping series.

1.2.3 Geometric Series

A geometric series has the form:

where is the first term and is the common ratio. The convergence behavior depends entirely on :

If : the series converges to
If : the series diverges

This is one of the most important series in mathematics, as many other series can be compared to or expressed in terms of geometric series.

1.2.4 The Harmonic Series and -Series

The harmonic series is:

Despite the terms approaching zero, this series diverges. This is a crucial counterexample showing that is necessary but not sufficient for convergence.

More generally, a -series has the form:

This series:

Converges if
Diverges if

The -series serves as a standard benchmark for comparison tests.

1.2.5 Necessary Condition for Convergence

The -th Term Test (or Divergence Test) states: If converges, then .

The contrapositive is more useful: If or the limit does not exist, then diverges.

Important: This test can only prove divergence. If , we cannot conclude anything about convergence (the harmonic series is the classic example).

1.3 Convergence Tests for Series with Non-Negative Terms

When for all , the partial sums form a non-decreasing sequence. Such series either converge (if the partial sums are bounded) or diverge to infinity (if unbounded). Several tests help determine which case applies.

1.3.1 Direct Comparison Test

Theorem (Direct Comparison Test): Let and be two series with for all (or at least for all sufficiently large ). Then:

If converges, then converges
If diverges, then diverges

Proof:

(1) Assume converges. Let and be the partial sums.

Since for all , we have:

The sequence is non-decreasing (since all terms are non-negative) and bounded above by (which exists since converges). By the Monotone Convergence Theorem, converges, so converges.

(2) This is the contrapositive of (1): if diverges to and , then the partial sums of must also diverge to .

Strategy: To use this test, find a series you already know converges or diverges (often a geometric series or -series) and compare term-by-term.

1.3.2 Limit Comparison Test

Theorem (Limit Comparison Test): Let and be two series with positive terms (, ). Define:

Then:

If , both series have the same behavior (both converge or both diverge)
If and converges, then converges
If and diverges, then diverges

Proof:

(1) Suppose . Then for any , there exists such that for all :

Choose . Then for :

This gives us:

If converges, then by the Direct Comparison Test (with the right inequality), converges.
If diverges, then by the Direct Comparison Test (with the left inequality), diverges.

(2) If , then for any , there exists such that for :

Thus . If converges and we choose , then converges by the Direct Comparison Test.

(3) If , then . By part (2) applied to and , if converges, then converges. Contrapositive: if diverges, then diverges.

Strategy: Choose to be a simpler series (geometric or -series) that resembles the dominant behavior of for large . The limit comparison test is often easier to apply than the direct comparison test since you don’t need an explicit inequality.

1.3.3 Ratio Test (D’Alembert’s Test)

Theorem (Ratio Test): Let be a series with positive terms, and let:

Then:

If , the series converges
If , the series diverges
If , the test is inconclusive

Proof:

(1) Suppose . Choose such that . Then there exists such that for all :

This gives us:

In general, for . Therefore:

Since this is a convergent geometric series (with ), the tail of converges, and thus converges.

(2) Suppose . Then there exists such that for all :

This means for all , so the terms are eventually increasing. In particular, does not approach 0, so by the -th term test, diverges.

(3) When , the test gives no information. Both convergent and divergent series can have (for example, diverges and converges, but both have ).

Application: The ratio test is particularly effective for series involving factorials or exponentials, where the ratio simplifies nicely.

1.3.4 Root Test (Cauchy’s Test)

Theorem (Root Test): Let be a series with non-negative terms, and let:

Then:

If , the series converges
If , the series diverges
If , the test is inconclusive

Proof:

(1) Suppose . Choose such that . Then there exists such that for all :

Thus for all . Since is a convergent geometric series (with ), by the Direct Comparison Test, converges.

(2) Suppose . Then there exists such that for all :

Thus for all , so . By the -th term test, diverges.

(3) When , the test is inconclusive (same reasoning as for the ratio test).

Application: The root test is particularly useful when involves -th powers, such as .

1.3.5 Comparison Between Ratio and Root Tests

Remark: There is an important relationship between the ratio and root tests:

In particular, if exists, then also exists and they are equal:

This means every series whose convergence can be determined by the ratio test can also be determined by the root test, but not vice versa (the root test is more powerful). However, the converse is not always true—some series can be analyzed by the root test but not by the ratio test.

In practice: Use the ratio test for series with factorials, and the root test for series with -th powers.

1.4 Alternating Series

An alternating series has the form:

where . The signs alternate between positive and negative.

1.4.1 Leibniz Test

Theorem (Leibniz Test): Let be an alternating series where . If:

The sequence is non-increasing (i.e., for all ), and

Then the alternating series converges.

Proof:

Let be the -th partial sum. We consider the even and odd partial sums separately.

Even partial sums: Consider :

Since , each parenthesis is non-negative, so:

Thus is non-decreasing.

We can also write:

Since each parenthesis is non-negative and :

So is bounded above. By the Monotone Convergence Theorem, converges to some limit .

Odd partial sums: We have:

Taking limits:

Since both subsequences converge to the same limit , the full sequence converges to .

Application: This test is particularly powerful because the conditions are easy to check, and it applies to many series that don’t meet the criteria for other tests.

1.5 Absolute and Conditional Convergence

1.5.1 Absolute Convergence

A series converges absolutely if the series of absolute values converges.

Theorem (The Absolute Convergence Test): If converges, then converges. In other words, absolute convergence implies convergence.

Proof:

We use the Cauchy criterion for convergence. Since converges, for any , there exists such that for all :

By the triangle inequality:

This shows that the partial sums of form a Cauchy sequence, and therefore converges.

Application: This is powerful because it allows us to apply tests designed for positive series (like the ratio test, root test, and comparison tests) to series with mixed signs by considering instead.

1.5.2 Conditional Convergence

A series is conditionally convergent if it converges, but does not converge absolutely. In other words, converges but diverges.

Example: The alternating harmonic series converges (by Leibniz test) to , but diverges. Hence it is conditionally convergent.

Important distinction: Absolutely convergent series have nice properties (they can be rearranged without changing the sum), while conditionally convergent series are more delicate (rearranging terms can change the sum or even make the series diverge).

1.5.3 Applying Tests for Absolute Convergence

To test for absolute convergence:

Ratio test for absolute convergence: Compute
Root test for absolute convergence: Compute

If , the series converges absolutely. If , the series diverges.

1.6 Power Series

1.6.1 Definition and Convergence

A power series centered at is a series of the form:

Unlike the series we’ve studied so far, a power series is a function of . The key question is: For which values of does the series converge?

1.6.2 The Convergence Theorem for Power Series

Theorem (The Convergence Theorem for Power Series): If a power series converges at a point , then it converges absolutely for all with . If the series diverges at , then it diverges for all with .

Proof:

Suppose converges. Then , so the sequence is bounded: there exists such that for all .

Now let be any value with . Then:

Since , the series is a convergent geometric series. By the Direct Comparison Test, converges, so converges absolutely.

The second statement follows by contrapositive: if converged for some , then by the first part, it would have to converge at , contradicting the assumption that it diverges there.

1.6.3 The Domain of Convergence of a Power Series

Theorem: For any power series , exactly one of the following holds:

The series converges only at (radius of convergence )
The series converges absolutely for all (radius of convergence )
There exists a finite such that:
- The series converges absolutely for
- The series diverges for
- At the endpoints , the series may either converge or diverge (must be checked separately)

The value is called the radius of convergence.

Proof:

Define:

This set is non-empty (it contains 0) and is an interval or or by the Convergence Theorem.

If , then .
If , then .
Otherwise, is finite and positive.

The set of all for which the series converges is called the interval of convergence. Depending on convergence at the endpoints, this can be any of: , , , or .

1.6.4 The Radius of Convergence of a Power Series

Theorem (Radius of Convergence by Ratio Test): For the power series , if the limit:

exists (or equals ), then the radius of convergence is given by:

Proof:

Apply the ratio test to the series :

The series converges absolutely when , i.e., .

Theorem (Hadamard Formula: Radius of Convergence by Root Test): For the power series , if the limit:

exists (or equals ), then the radius of convergence is given by the same formulas as in the ratio test.

Proof:

Apply the root test to :

The series converges absolutely when .

Important: These formulas give the radius. To find the interval of convergence, you must separately check convergence at the endpoints and .

1.6.5 Operations on Power Series

Theorem (Operations on Power Series):

Addition and Subtraction: If has radius and has radius , then has radius at least .
Scalar Multiplication: If has radius , then has the same radius .
Cauchy Product of Power Series: If and both have positive radius of convergence, then their product is:

The product series has radius at least where and are the radii of and .

1.7 Properties of Absolutely Convergent Series

1.7.1 Rearrangement Theorem

Definition: A rearrangement of a series is a series where for some bijection (i.e., the terms are the same but in a different order).

Theorem (The Rearrangement Theorem for Absolutely Convergent Series): If converges absolutely to a sum , and is any rearrangement of , then also converges absolutely to .

Proof:

Since converges absolutely, converges. Let and .

Let . Since converges, there exists such that:

Let be large enough so that (i.e., the first terms of the original series all appear among the first terms of the rearranged series).

For , the partial sum differs from by at most terms from , so:

Therefore, , proving that the rearranged series converges to the same sum.

Remark: In contrast, conditionally convergent series do not have this property. The Riemann Rearrangement Theorem states that if converges conditionally, then for any extended real number , there exists a rearrangement of that converges to (or diverges if )!

1.7.2 Cauchy Product

Theorem: If and are two absolutely convergent series with sums and respectively, then their Cauchy product:

converges absolutely to .

Theorem (Merten’s Theorem): Suppose and are two convergent series, and at least one of them converges absolutely. Then their Cauchy product converges to , where:

Proof (outline):

Without loss of generality, assume converges absolutely. Let , , and be the partial sums.

We can write:

We want to show that .

Consider:

Since , we have as . Using the absolute convergence of and careful estimates, we can show:

Since and is constant:

Remark: If both series are only conditionally convergent (neither converges absolutely), their Cauchy product may diverge. For example, the Cauchy product of with itself (which converges conditionally by Leibniz) results in a divergent series.

1.8 Asymptotic Notation: Big O, little o, and Asymptotic Equivalence

Asymptotic notation provides a convenient way to compare the behavior of sequences or functions as (or as ).

1.8.1 Big O Notation

Definition: We write (read “ is big O of ”) if there exists a constant and such that:

This means grows no faster than (up to a constant multiple).

Example: because for large , the term dominates.

1.8.2 Little o Notation

Definition: We write (read “ is little o of ”) if:

This means grows much slower than .

Example: because .

1.8.3 Asymptotic Equivalence

Definition: We write (read “ is asymptotically equivalent to ” or “ is asymptotic to ”) if:

This means and have the same asymptotic behavior.

Example: because .

1.8.4 Properties and Applications

Key Properties:

If , then
If , then
If and converges, then converges
If and diverges, then diverges

Application to Series: When analyzing series, we can often replace complicated terms with simpler asymptotically equivalent expressions.

Example: Consider .

For large :

Numerator: (exponentials dominate polynomials)
Denominator: (exponentials dominate logarithms)

Therefore:

Since is a convergent geometric series (with ratio ), the original series also converges.

2. Definitions

Sequence: An ordered list of numbers where each term is determined by a rule or formula.
Limit of a Sequence: means that the terms get arbitrarily close to as increases.
Convergent Sequence: A sequence that has a finite limit.
Divergent Sequence: A sequence that does not converge to a finite limit.
Bounded Sequence: A sequence is bounded if there exist numbers and such that for all .
Monotonic Increasing Sequence: A sequence where for all .
Monotonic Decreasing Sequence: A sequence where for all .
Bolzano-Weierstrass Theorem: Every bounded monotonic sequence converges.
Squeeze Theorem: If for large and , then .
Series: The sum of infinitely many terms from a sequence, denoted .
Partial Sum: The -th partial sum is .
Convergent Series: A series whose sequence of partial sums converges to a finite limit , called the sum of the series.
Divergent Series: A series whose sequence of partial sums does not converge to a finite limit.
Telescoping Series: A series of the form where consecutive terms cancel.
Geometric Series: A series of the form where is the first term and is the common ratio.
Harmonic Series: The series , which diverges.
-Series: A series of the form , which converges if and diverges if .
Alternating Series: A series with terms that alternate in sign, of the form where .
Absolutely Convergent Series: A series for which converges.
Conditionally Convergent Series: A series that converges but does not converge absolutely.
Power Series: A series of the form where is a variable.
Radius of Convergence: The value such that a power series converges absolutely for and diverges for .
Interval of Convergence: The set of all values of for which a power series converges.
Cauchy Product: For two series and , the Cauchy product is where .
Big O Notation (): We write if there exists and such that for all .
Little o Notation (): We write if .
Asymptotic Equivalence (): We write if .

3. Formulas

Limit Definition of :
Generalized Limit:
Squeeze Theorem: If and , then .
Recursive Limit Evaluation: If and where is continuous, then .
-th Partial Sum:
Series Convergence: if .
Telescoping Series Sum: If , then
Geometric Series Sum: For ,
Mengoli Series:
Generalized Telescoping Sum:
Harmonic Series Divergence:
-Series Convergence: converges if and only if .
Linear Property of Series:
-th Term Test: If converges, then .
Direct Comparison Test: If , then: (1) converges converges; (2) diverges diverges.
Limit Comparison Test: If where , then and have the same behavior.
Ratio Test: Let . If , the series converges absolutely; if , the series diverges; if , the test is inconclusive.
Root Test: Let . If , the series converges absolutely; if , the series diverges; if , the test is inconclusive.
Leibniz Test: If is non-increasing and , then converges.
Absolute Convergence Test: If converges, then converges.
Alternating Harmonic Series Sum:
Radius of Convergence (Ratio Test):
Radius of Convergence (Root Test):
Cauchy Product: where

4. Practice

4.1. Find the Limit with Factorials and Powers (Lab 9, Task 1)

Find the limit:

Click to see the solution

Key Concept: Simplify factorials and manipulate the expression to match the standard form .

Simplify the factorial ratio:
Combine the powers:
Rewrite the numerator:
Express in standard form:
Manipulate the exponent: Let . We want to rewrite this in the form :
Evaluate the limit: As , we have , so:

Also:

Therefore:

Answer:

4.2. Convergence of Geometric Series (a) (Lab 9, Task 1a)

Determine the convergence or divergence of the following geometric series:

Click to see the solution

Key Concept: A geometric series converges if and its sum is .

Identify the first term and common ratio:
- First term: (when , we get )
- Common ratio:
Check the convergence condition:

Therefore, the series converges.
Calculate the sum:

Answer: The series converges to .

4.3. Convergence of Geometric Series (b) (Lab 9, Task 1b)

Determine the convergence or divergence of the following geometric series:

Click to see the solution

Identify the first term and common ratio:
- First term:
- Common ratio:
Check the convergence condition:

Therefore, the series converges.
Calculate the sum:

Answer: The series converges to .

4.4. Convergence of Geometric Series (c) (Lab 9, Task 1c)

Determine the convergence or divergence of the following geometric series:

Click to see the solution

Identify the first term and common ratio:
- First term:
- Common ratio:
Check the convergence condition:

Therefore, the series diverges.

Answer: The series diverges.

4.5. Limit with Exponential and Rational Terms (Lab 9, Task 2)

Find the limit:

Click to see the solution

Key Concept: Evaluate each term separately, recognizing that exponentials with base less than 1 approach 0, and transform the second term to match the limit form.

Evaluate the first term:

(since , the powers approach 0)
Rewrite the denominator of the second term:
Transform to standard form:
Evaluate the limit: As , we have , so:

Therefore:

And:
Evaluate the second term:
Combine both terms:

Answer:

4.6. Parametric Convergence of Geometric Series (a) (Lab 9, Task 2a)

For which values of does the following series converge, and what is its sum?

Click to see the solution

Key Concept: Rewrite as a geometric series and find when .

Rewrite as a geometric series:

This is a geometric series with first term and common ratio .
Find the convergence condition: The series converges when :

This is true for all since and .
Calculate the sum:

Answer: The series converges for all , and its sum is .

4.7. Parametric Convergence of Geometric Series (b) (Lab 9, Task 2b)

For which values of does the following series converge, and what is its sum?

Click to see the solution

Rewrite as a geometric series:

This is a geometric series with and .
Find the convergence condition: The series converges when :
Analyze both inequalities:
- Right inequality: This is true for all .
- Left inequality: This is true for all .
Combine conditions: The series converges for all .
Calculate the sum:

Answer: The series converges for all , and its sum is .

4.8. Limit with Factorial, Exponential, and Oscillating Terms (Lab 9, Task 3)

Find the limit:

Click to see the solution

Key Concept: Use the squeeze theorem for the oscillating term and recognize that factorials grow faster than exponentials.

Evaluate the first term: Factorials grow much faster than exponentials, so:

(This can be proven using the ratio test: )
Evaluate the second term’s power:
Handle the oscillating factor using the squeeze theorem: We have , so dividing by :
Apply limits to the bounds:
Conclude by the squeeze theorem:
Evaluate the second term:
Combine both terms:

Answer:

4.9. Ratio Test (a) (Lab 9, Task 3a)

Determine the convergence of the following series using the ratio test:

Click to see the solution

Key Concept: For series with factorials, the ratio test is usually effective.

Set up the ratio:
Simplify:
Apply the ratio test: Since , the series converges (absolutely).

Answer: The series converges.

4.10. Ratio Test (b) (Lab 9, Task 3b)

Determine the convergence of the following series using the ratio test:

Click to see the solution

Set up the ratio:
Simplify:
Apply the ratio test: Since , the series converges.

Answer: The series converges.

4.11. Convergence of a Recursive Sequence (Lab 9, Task 4)

Let be a sequence such that and for .

(a) Prove by induction that the sequence is bounded below by 1, i.e., for all .

(b) Prove that the sequence is decreasing, i.e., for all .

(c) Conclude that the sequence converges and find its limit.

Click to see the solution

Key Concept: Use induction to establish boundedness, analyze the difference for monotonicity, then apply the Bolzano-Weierstrass theorem.

(a) Prove by induction:

Base case ():
Inductive hypothesis: Assume for some .
Inductive step: We need to show . From the recursive formula:

We want to show:

This is true since by the inductive hypothesis.

Therefore, for all .

(b) Prove the sequence is decreasing:

Compute the difference:
Find a common denominator:
Analyze the sign: The denominator since . For the numerator:
- If , then , so
- If , then
- If , then , so
Determine the behavior: Since and , we can verify that once , we have . The sequence decreases from toward .

Therefore, the sequence is decreasing (at least after the first term and while ).

(c) Find the limit:

Apply Bolzano-Weierstrass theorem: The sequence is bounded below (by 1) and decreasing, so it converges.
Let : Taking limits on both sides of the recursive formula:
Solve for :
Choose the correct value: Since for all , we must have . Therefore, .

Answer:

(a) The sequence is bounded below by 1 (proven by induction).

(b) The sequence is decreasing (proven by analyzing the sign of ).

(c) The sequence converges to .

4.12. Root Test (a) (Lab 9, Task 4a)

Determine the convergence of the following series using the root test:

Click to see the solution

Key Concept: The root test is ideal for series where the general term involves -th powers.

Compute the -th root:
Simplify:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.13. Root Test (b) (Lab 9, Task 4b)

Determine the convergence of the following series using the root test:

Click to see the solution

Compute the -th root:
Simplify using exponent rules:
Rewrite the base:

Therefore:
Evaluate the limit:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.14. Telescoping Series with Partial Fractions (a) (Lab 10, Part I, Example 1)

Find the sum of the series:

Click to see the solution

Key Concept: Use partial fraction decomposition to create a telescoping series.

Decompose using partial fractions:
Solve for and :

Setting :

Setting :
Rewrite the series:
Compute the partial sum:

Rearranging:
Take the limit:

Answer:

4.15. Telescoping Series with Square Roots (a) (Lab 10, Part I, Homework H1)

Determine whether the series converges and find its sum if possible:

Click to see the solution

Key Concept: Recognize that this is a second-order telescoping series.

Rewrite the general term: Let . Notice this can be written as:
Define differences: Let . Then:
This is a telescoping series of differences:
Evaluate the limit:

Therefore:

Answer: The series converges to

4.16. -th Term Test Example (a) (Lab 10, Part II, Example 1)

Study the convergence of:

Click to see the solution

Key Concept: Transform the expression to recognize the limit involving .

Rewrite the general term:
Find the limit:

Let . As , we have , and:

As :
Apply the -th term test: Since , the series diverges.

Answer: The series diverges.

4.17. Asymptotic Equivalence Example (a) (Lab 10, Part II, Homework H1)

Study the convergence of:

Click to see the solution

Key Concept: Use the asymptotic behavior as .

Apply asymptotic equivalence: For large , is small, so:

Therefore:
Find the limit:

(using L’Hôpital’s rule or the fact that )
Apply the -th term test: Since , the series diverges.

Answer: The series diverges.

4.18. Direct Comparison Test with Logarithm (Lab 10, Part III, Example 1)

Study the convergence of:

Click to see the solution

Key Concept: Simplify the logarithm and compare with a -series.

Simplify the logarithm: For large , , so:
Find the asymptotic behavior: More precisely:

Since as :
Compare with the harmonic series:

Since diverges, by the limit comparison test, also diverges.

Answer: The series diverges.

4.19. -Series Example (Lab 10, Part III, Homework H1)

Study the convergence of:

Click to see the solution

Key Concept: Recognize this as a -series.

Identify the -series: This is a -series with :
Apply the -series test: A -series converges if and only if .

Here, , so the series diverges.

Answer: The series diverges.

4.20. Ratio Test with Factorial (a) (Lab 10, Part IV, Example 1)

Study the convergence of:

Click to see the solution

Key Concept: For series involving factorials, the ratio test is usually effective.

Apply the ratio test:
Simplify:
Apply the ratio test: Since , the series diverges.

Answer: The series diverges.

4.21. Ratio Test with Powers and Factorials (a) (Lab 10, Part IV, Homework H1)

Study the convergence of:

Click to see the solution

Apply the ratio test:
Simplify:
Apply the ratio test: Since , the series converges.

Answer: The series converges.

4.22. Root Test with Double Exponential (Lab 10, Part V, Example 1)

Study the convergence of:

Click to see the solution

Key Concept: The root test is ideal when the general term has -th powers.

Apply the root test:
Simplify:

(since exponentials grow much faster than polynomials)
Apply the root test: Since , the series converges.

Answer: The series converges.

4.23. Root Test with Exponential Powers (a) (Lab 10, Part V, Homework H1)

Study the convergence of:

Click to see the solution

Apply the root test:
Simplify:
Evaluate limits:
- (taking log: )
Combine:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.24. Series with Factorial and Odd Numbers (Lab 10, Part I, Example 2)

Find the sum of the series:

Click to see the solution

Key Concept: Express the general term as a difference to create a telescoping series.

Denote the product of odd numbers: Let .
Find a telescoping form: Notice that:

Since , we have , so:

Therefore:
Write the partial sum:

This telescopes:

(where )
Take the limit:

Answer:

4.25. Series with Factorial in Denominator (Lab 10, Part I, Homework H2)

Determine whether the series converges and find its sum if possible:

Click to see the solution

Key Concept: Break the series into simpler parts and use known series involving factorials.

Split the series:
Evaluate the second sum: The Taylor series for gives us:
Evaluate the first sum: For , write :

For the first part (note terms with contribute 0):

For the second part (note term with contributes 0):

Therefore:
Combine the results:

Answer: The series converges to

4.26. -th Term Test Example (b) (Lab 10, Part II, Example 2)

Study the convergence of:

Click to see the solution

Find the limit of the general term:
Apply the -th term test: Since , the series diverges by the -th term test.

Answer: The series diverges.

4.27. Comparison Test Example (b) (Lab 10, Part II, Homework H2)

Study the convergence of:

Click to see the solution

Key Concept: Rationalize the denominator to find an equivalent form.

Rationalize the denominator:
Analyze the term:
Compare with a known series: Since and diverges (it’s a -series with , or equivalently, the terms grow without bound), the series diverges.

Answer: The series diverges.

4.28. Direct Comparison Test with Geometric Series (Lab 10, Part III, Example 2)

Study the convergence of:

Click to see the solution

Key Concept: For series with exponentials in the denominator, compare with a geometric series.

Bound the numerator: For all , we have , so:
Apply the ratio test: Let . Then:

Therefore, converges, and so does .
Apply the comparison test: By the direct comparison test, also converges.

Answer: The series converges.

4.29. Integral Test or Comparison (Lab 10, Part III, Homework H2)

Study the convergence of:

Click to see the solution

Key Concept: Use the integral test or comparison with known series.

Apply the integral test: Consider the function for . This function is positive and decreasing for .

Evaluate the improper integral:
Use substitution: Let , so :
Conclude: Since the integral diverges, by the integral test, the series diverges.

Answer: The series diverges.

4.30. Ratio Test with Exponential (b) (Lab 10, Part IV, Example 2)

Study the convergence of:

Click to see the solution

Apply the ratio test:
Simplify:
Apply the ratio test: Since , the series converges.

Answer: The series converges.

4.31. Ratio Test with Stirling’s Approximation (Lab 10, Part IV, Homework H2)

Study the convergence of:

Click to see the solution

Apply the ratio test:
Simplify:
Apply the ratio test: Since , the series converges.

Answer: The series converges.

4.32. Root Test with Rational Powers (Lab 10, Part V, Example 2)

Study the convergence of:

Click to see the solution

Apply the root test:
Simplify:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.33. Root Test with Pattern (Lab 10, Part V, Homework H2)

Study the convergence of:

Click to see the solution

Key Concept: Identify the pattern and express the general term.

Identify the pattern: Looking at the terms:
- :
- :
- :
The general term is:
Apply the root test: This is challenging directly. Instead, note that:
Find the limit:

As :
Therefore:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.34. Series with Trigonometric Functions (Lab 10, Part I, Example 3)

Find the sum of the series:

Click to see the solution

Key Concept: Use the trigonometric identity to create a telescoping series.

Apply the sine difference formula: Note that:
Recognize the pattern: Since , and for small , , we can use:

This follows from the tangent difference formula.
Write the partial sum:

This telescopes:
Take the limit:

Answer:

4.35. Series with Oscillating Sine Function (Lab 10, Part II, Example 3)

Study the convergence of:

Click to see the solution

Key Concept: The terms oscillate and can be very small, making very large.

Analyze the general term: The values for are dense in and can get arbitrarily close to 0. When is close to 0, becomes arbitrarily large.
Consider the limit: The sequence does not have a limit as (it oscillates), but more importantly, there are infinitely many terms that are very large.
Comparison argument: By Dirichlet’s approximation theorem, there exist infinitely many integers such that for some integer . For such , for some constant , giving , which would make the series diverge.

Answer: The series diverges (the terms do not approach zero consistently, and infinitely many terms are unbounded).

4.36. Limit Comparison Test with -Series (Lab 10, Part III, Example 3)

Study the convergence of:

Click to see the solution

Key Concept: Identify the dominant behavior for large and use limit comparison.

Determine asymptotic behavior: For large :
- Numerator:
- Denominator:
Therefore:
Apply limit comparison test: Let and .

Dividing numerator and denominator by :
Conclude: Since and converges (-series with ), the original series also converges.

Answer: The series converges.

4.37. Logarithmic Series (Lab 10, Part III, Homework H3)

Study the convergence of:

Click to see the solution

Key Concept: Use the root test or comparison with exponential decay.

Apply the root test:
Evaluate the exponent:

Therefore:
Root test is inconclusive: Since , the root test is inconclusive.
Alternative approach - comparison: For large , , so .

Therefore:

Since , and we need for convergence, this comparison doesn’t immediately help. However, the term grows much faster than any polynomial, so the series converges rapidly.
More precise analysis: Since grows faster than for any fixed , we have:

for sufficiently large . By comparison with the convergent -series , the series converges.

Answer: The series converges.

4.38. Comparison Test Analysis (Lab 10, Part III, Example 4)

Study the convergence of:

Click to see the solution

Simplify the expression:
Compare with harmonic series: The series becomes:

This is the harmonic series starting from , which diverges.

Answer: The series diverges.

4.39. Epsilon-Delta Definition: Limit of Square Root (Lab 11, Task 1)

Prove by epsilon-delta definition that

Click to see the solution

Key Concept: Use the definition of a limit to find a suitable for any given such that whenever .

Set up the inequality: For any given , we need to find such that:
Manipulate the inequality: For all :
Find : We need the interval centered at to be inside .

The distance from to the endpoints is:
- Left endpoint:
- Right endpoint:
For , take:
For : Take

For example:
- If :
- If :
Conclusion: With this choice of , the inequality automatically places in the desired interval, making .

Answer: For any , choose when , or when . This proves .

4.40. Epsilon-Delta Definition: Piecewise Function (Lab 11, Task 2)

Let Prove by epsilon-delta definition that .

Click to see the solution

Key Concept: The limit depends on the behavior near , not at . Since for all , we work with .

Set up the inequality: For any given , we need to find such that:
Work with the inequality (assuming ): For , we have :
Find for : We want the interval centered at to lie inside .

Take to be the distance from to the nearer endpoint:
For : Take:
Conclusion: With this choice of , the inequality ensures is in the interval where .

Answer: For any , choose when , or when . This proves .

4.41. Leibniz Test Application (a) (Lab 11, Task 3a)

Use the Leibniz test to show that the following series converges:

Click to see the solution

Key Concept: Apply the Leibniz test by checking that the terms are positive, decreasing, and approach zero.

Verify positivity:
Verify decreasing:
Verify limit is zero:
Apply Leibniz test: All three conditions are satisfied, so the series converges.

Answer: The series converges by the Leibniz test. (It converges to .)

4.42. Leibniz Test Application (b) (Lab 11, Task 3b)

Use the Leibniz test to show that the following series converges:

Click to see the solution

Verify positivity:
Verify decreasing:
Verify limit is zero:
Apply Leibniz test: All conditions of the alternating series test are satisfied.

Answer: The series converges by the Leibniz test.

4.43. Leibniz Test Application (c) (Lab 11, Task 3c)

Use the Leibniz test to show that the following series converges:

Click to see the solution

Verify positivity:
Verify decreasing: Check if by comparing the ratio:

For : , so , thus . ✓
Verify limit is zero:
Apply Leibniz test: All conditions are satisfied.

Answer: The series converges by the Leibniz test.

4.44. Leibniz Test Application (d) (Lab 11, Task 3d)

Use the Leibniz test to show that the following series converges:

Click to see the solution

Verify positivity:
Verify decreasing: Check the ratio:

For : (equality when )

Thus for all . ✓
Verify limit is zero:

(Exponentials grow faster than polynomials) ✓
Apply Leibniz test: All conditions are satisfied.

Answer: The series converges by the Leibniz test.

4.45. Absolute Convergence Classification (a) (Lab 11, Task 4a)

Classify the series as absolutely convergent, conditionally convergent, or divergent:

Click to see the solution

Key Concept: Test for absolute convergence first using the ratio test.

Test for absolute convergence: Consider:
Apply the ratio test:
Conclude: Since , the series of absolute values converges.

Answer: The series converges absolutely.

4.46. Absolute Convergence Classification (b) (Lab 11, Task 4b)

Classify the series as absolutely convergent, conditionally convergent, or divergent:

Click to see the solution

Test for absolute convergence: Consider:
Use comparison test: Since for all :
Compare with -series: The series is a convergent -series with .
Apply comparison test: By the comparison test, converges.

Answer: The series converges absolutely.

4.47. Absolute Convergence Classification (c) (Lab 11, Task 4c)

Classify the series as absolutely convergent, conditionally convergent, or divergent:

Click to see the solution

Key Concept: Check absolute convergence first, then conditional convergence if absolute convergence fails.

Test for absolute convergence: Consider:
Use comparison with harmonic series: For : , so:

Since diverges (harmonic series), by comparison test, also diverges.
The series does NOT converge absolutely.
Test for conditional convergence using Leibniz test: Let .
- Positivity: for all ✓
- Limit is zero: Using L’Hôpital’s rule:
- Decreasing: We need to verify . Consider:
  
  As , this approaches , so eventually . ✓
Apply Leibniz test: The series converges by the Leibniz test.

Answer: The series converges conditionally.

4.48. Absolute Convergence Classification (d) (Lab 11, Task 4d)

Classify the series as absolutely convergent, conditionally convergent, or divergent:

Click to see the solution

Key Concept: Note that , so this is an alternating series.

Rewrite the series:
Test for absolute convergence: Consider:
Use limit comparison test with :

Since diverges and the limit is positive and finite, also diverges.
The series does NOT converge absolutely.
Test for conditional convergence using Leibniz test: Let .
- Positivity: for all ✓
- Limit is zero:
- Decreasing: Check the ratio:
  
  We need to show this is less than 1:
  
  So . ✓
Apply Leibniz test: The series converges by the Leibniz test.

Answer: The series converges conditionally.

4.49. Power Series: Radius and Interval (a) (Lab 11, Task 5a)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints.

Click to see the solution

Key Concept: Use the ratio test to find the radius of convergence, then check endpoints separately.

Apply the ratio test: Let . Then:
Simplify:
Find the radius: The series converges absolutely when , i.e., .

Radius of convergence:
Check the endpoint :

By Stirling’s approximation or induction, we can show that for all .

Since the terms don’t approach zero, the series diverges.
Check the endpoint :

Since , the terms don’t approach zero, so the series diverges.

Answer: Radius of convergence . Interval of convergence: .

4.50. Power Series: Radius and Interval (b) (Lab 11, Task 5b)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints.

Click to see the solution

Apply the ratio test (for ):
Find the radius: The series converges when , i.e., .

Radius of convergence:
Check the endpoints:
- At : Diverges.
- At : Since , the series diverges.

Answer: Radius of convergence . Interval of convergence: . Absolute convergence occurs in .

4.51. Power Series: Radius and Interval (c) (Lab 11, Task 5c)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints.

Click to see the solution

Apply the ratio test:
Evaluate the limit:

So .
Find the radius: The series converges absolutely when , i.e., .

Radius of convergence:
Check the endpoint :

Since for , we have .

The harmonic series diverges, so by comparison, diverges.
Check the endpoint :

Check the Alternating Series Test conditions:
- for ✓
- is decreasing ✓
- ✓
By the Leibniz test, the series converges.

Answer: Radius of convergence . Interval of convergence: . The series converges conditionally at .

4.52. Power Series: Radius and Interval (d) (Lab 11, Task 5d)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints.

Click to see the solution

Apply the ratio test:
Find the radius: The series converges absolutely when , i.e., .

Radius of convergence:
Check the endpoint :

Check the Alternating Series Test:
- for all ✓
- is decreasing ✓
- ✓
All conditions are satisfied, so the series converges by the Leibniz test.
Check the endpoint :

This is a negative -series with , which diverges.

Answer: Radius of convergence . Interval of convergence: . The series converges conditionally at .

4.53. Direct Comparison Test: Series Less Than Geometric (Chapter 4, Example 1)

Determine the convergence of the series .

Click to see the solution

Key Concept: Find a larger convergent series to compare with.

Find an upper bound: For , we have . To see this:
- For : ✓
- For : ✓
- For : ✓
Therefore:
Use the comparison test: The series is a geometric series with , so it converges.

By the direct comparison test, since and converges, we conclude that converges.

Answer: The series converges.

4.54. Limit Comparison Test: Convergent Series (Chapter 4, Example 2)

Determine the convergence of the series .

Click to see the solution

Key Concept: For large , the term behaves like , so compare with the -series .

Choose a comparison series: Let and .
Compute the limit of the ratio:
Expand and simplify:

Therefore:
Apply the limit comparison test: Since and converges (it’s a -series with ), we conclude that also converges.

Answer: The series converges.

4.55. Ratio Test: Convergent Series with Factorial (Chapter 4, Example 3)

Determine the convergence of the series .

Click to see the solution

Compute the ratio:
Simplify:
Apply the ratio test: Since , the series converges.

Answer: The series converges.

4.56. Root Test: Convergent Series (Chapter 4, Example 4)

Determine the convergence of the series .

Click to see the solution

Compute the -th root:
Simplify:
Evaluate the limit of : Taking logarithms: as . Therefore, .
Complete the calculation:
Apply the root test: Since , the series converges.

Answer: The series converges.

4.57. Absolute vs. Conditional Convergence (a) (Chapter 4, Example 5)

Determine whether the series converges absolutely, conditionally, or diverges.

Click to see the solution

Key Concept: Check absolute convergence first, then use the Leibniz test if needed.

Test for absolute convergence: Consider the series of absolute values:

This is a -series with , so it converges.
Conclusion: Since converges, the original series converges absolutely.

Answer: The series converges absolutely.

4.58. The Series of Mengoli (Chapter 4, Example 6)

For the series (the series of Mengoli), find the sum.

Click to see the solution

Key Concept: Use partial fraction decomposition to create a telescoping series.

Decompose using partial fractions:
Solve for and :

Comparing coefficients:
- Coefficient of :
- Constant term:
Therefore, and .
Rewrite the series:
Compute the -th partial sum:

Most terms cancel (telescope):
Take the limit:

Answer: The series converges to .

4.59. Divergence of a Logarithmic Telescoping Series (Chapter 4, Example 7)

Show that the series diverges.

Click to see the solution

Key Concept: Use logarithm properties to create a telescoping series, then evaluate the limit of partial sums.

Rewrite using logarithm properties:
This is a telescoping series:
Compute the -th partial sum:

(since )
Evaluate the limit:

Answer: The series diverges to .

4.60. Direct Comparison Test: Divergent Series (Chapter 4, Example 8)

Show that the series diverges.

Click to see the solution

Key Concept: Find a smaller divergent series to compare with.

Find a lower bound: Note that for all . Therefore:

(Alternatively, recognize this as a -series with )
Use the comparison test: The harmonic series diverges.

By the direct comparison test, since and diverges, we conclude that also diverges.

Answer: The series diverges (it’s a -series with ).

4.61. Limit Comparison Test: Divergent Series (Chapter 4, Example 9)

Determine the convergence of the series .

Click to see the solution

Choose a comparison series: For large , the term behaves like . Let and .
Compute the limit of the ratio:
Apply the limit comparison test: Since and diverges (harmonic series), we conclude that also diverges.

Answer: The series diverges.

4.62. Ratio Test: Divergent Series with Factorial (Chapter 4, Example 10)

Determine the convergence of the series .

Click to see the solution

Compute the ratio:
Simplify:

(using the result from the previous problem)
Apply the ratio test: Since , the series diverges.

Answer: The series diverges.

4.63. Root Test: Divergent Series (Chapter 4, Example 11)

Determine the convergence of the series .

Click to see the solution

Compute the -th root:
Simplify using exponent rules:
Evaluate the limit:
Apply the root test: Since , the series diverges.

Answer: The series diverges.

4.64. Absolute vs. Conditional Convergence (b) (Chapter 4, Example 12)

Determine whether the series converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence:

This is the harmonic series, which diverges. So the series does not converge absolutely.
Test for convergence using Leibniz test: The series has the form where .
- Non-increasing: ✓
- Limit is zero: ✓
By the Leibniz test, the series converges.
Conclusion: The series converges but not absolutely, so it is conditionally convergent.

Answer: The series converges conditionally.

4.65. Absolutely Convergent Series with Ratio Test (Chapter 4, Example 13)

Determine whether the series converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence using the ratio test: Let (ignoring the ).
Simplify:
Apply the ratio test: Since , the series of absolute values converges.

Answer: The series converges absolutely.

4.66. Conditionally Convergent Series (Chapter 4, Example 14)

Determine whether the series converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence: Consider:
Use limit comparison test: For large , the dominant terms are:

Let and .
Conclude about absolute convergence: Since and diverges (p-series with ), the series does not converge absolutely.
Test for convergence using Leibniz test: Let .
- Limit is zero:
- Non-increasing: This requires verification, but for large : which is decreasing. (A rigorous proof would show for large .)
Apply Leibniz test: Both conditions are satisfied, so the series converges.

Answer: The series converges conditionally.

4.67. Alternating Harmonic Series (Chapter 4, Example 15)

Show that the series converges, and state its sum.

Click to see the solution

Key Concept: Apply the Leibniz test for alternating series.

Identify the alternating series:

This has the form where .
Check the conditions of the Leibniz test:
- Non-increasing: Is ?
- Limit is zero: Does ?
Apply the Leibniz test: Both conditions are satisfied, so the series converges.
State the sum: It can be shown (using advanced techniques) that:

Answer: The series converges to .

4.68. Absolute Convergence with Parameter (Chapter 4, Example 16)

For which values of does the series converge absolutely, conditionally, or diverge?

Click to see the solution

Key Concept: Use the ratio test for absolute convergence, treating as a parameter.

Test for absolute convergence: Consider:
Apply the ratio test:
Determine absolute convergence:
- If , i.e., or , the series converges absolutely.
- If , i.e., , the series diverges.
- If , i.e., , the test is inconclusive.
Check the boundary cases :
- When : This is an alternating series with non-increasing and approaching 0. By Leibniz, it converges. Since the absolute version is which diverges (comparable to harmonic series), it converges conditionally.
- When : Since : By the same reasoning as , this converges conditionally.

Answer:

Absolutely convergent for
Conditionally convergent for
Divergent for

4.69. Find Sum of Series (a) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Method: Telescoping Series

Simplify the general term: Using partial fraction decomposition:
Write the partial sum : Expanding the terms allows us to cancel the middle terms:
Identify remaining terms: The positive terms and remain at the beginning. The negative terms and remain at the end.
Calculate the limit:

Answer:

4.70. Find Sum of Series (b) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Decompose into partial fractions: The difference between factors is .
Analyze the terms: This is a telescoping series with a “gap”. The terms of the denominator form the arithmetic progression: The subtraction removes terms starting from the 4th term of the sequence ().
Determine survivors: The first three positive terms will not be cancelled:
- :
- :
- :
All terms at the “infinity end” approach 0.
Calculate the sum:

Answer:

4.71. Find Sum of Series (c) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Decompose: The difference in the denominator is 1.
Telescope: Only the very first term remains, as the last term tends to 0.
Result:

Answer:

4.72. Find Sum of Geometric-Like Series (d) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Write the partial sum:
Cancel and take the limit: All intermediate terms cancel.

Answer:

4.73. Find Sum with Arctangent (e) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Identify the telescoping form: Let . The series is .
Compute partial sum limit:
Take the limit: As , .

Answer:

4.74. Geometric Series (a) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Identify parameters: This is a geometric series or similar.
- First term ():
- Common ratio:
Calculate sum: Since , the sum is valid.

Answer:

4.75. Geometric Series (b) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Split the series:
Sum each geometric series:
- First:
- Second:
Combine:

Answer:

4.76. Geometric Series (c) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Identify parameters: rewrite as .
- First term:
- Ratio:
Calculate sum:

Answer:

4.77. Geometric Series (d) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Isolate -th power:
Identify parameters:
- First term (of the sum part):
- Ratio: .
Calculate:

Answer:

4.78. Geometric Series (e) (Practice Sheet, Task 1)

Find the sum of the following series:

Click to see the solution

Identify parameters: This is .
- First term ():
- Ratio: . Since , .
Calculate sum:

Answer: