LESSON X



LESSON 20 POSITIVE TERM SERIES

Definition Given the series [pic], if [pic] for all [pic], then [pic] is called a positive term series.

Recall the following from MATH-1850. Let a function f be defined on an interval [pic] and let [pic] and [pic] be any numbers in the interval. Then f is decreasing on the interval if [pic] whenever [pic].

Definition A sequence [pic], with a domain of [pic], is decreasing on this domain if [pic] for all [pic].

Theorem (The Integral Test) Given the positive term series [pic], let [pic]. If the function f is continuous and decreasing on the closed interval [pic], then

1. the series [pic] converges if the improper integral [pic] converges, and

2. the series [pic] diverges if the improper integral [pic] diverges.

Proof Will be provided later.

Examples Determine whether the following series converge or diverge using the Integral Test if applicable.

1. [pic]

Let [pic]

The function f is positive on the closed interval [pic]. Now, we just need to check that f is decreasing on the interval [pic].

[pic] [pic]

[pic], which is not in the closed interval [pic]

[pic] is defined for all real numbers

Sign of [pic]: [pic]

(

1

Since [pic] for all x in the interval [pic], then the function f is decreasing on the closed interval [pic].

Thus, we can apply the Integral Test.

We need to evaluate the improper integral [pic].

[pic] = [pic]

The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.

Using the integral formula [pic], where [pic], we have that [pic]. Thus,

[pic] = [pic] = [pic] =

[pic] = [pic] since

[pic]

Thus, the improper integral [pic] converges. Thus, by the Integral Test, the series [pic] converges.

Answer: Converges; Integral Test

2. [pic]

Since [pic] = [pic], then we only need to check to whether the series [pic] converges or diverges.

Let [pic]

The function f is positive on the closed interval [pic]. Now, we just need to check that f is decreasing on the interval [pic].

[pic] =

[pic] = [pic]

[pic], which is not in the closed interval [pic], and [pic], which is in the interval [pic]

[pic] is undefined for [pic], which is not in the interval [pic]

Sign of [pic]: + [pic]

( (

1 [pic]

NOTE: [pic].

The function f is not decreasing on the closed interval [pic]. However, since [pic] for all x in the open interval [pic] and [pic], then the function f is decreasing on the closed interval [pic].

Thus, we can apply the Integral Test to the series [pic].

Since [pic] = [pic], then if the series

[pic] converges, then the series [pic] converges.

We need to evaluate the improper integral [pic].

[pic] = [pic]

The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.

[pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic] = [pic]

Thus,

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

Thus, the improper integral [pic] converges. Thus, by the

Integral Test, the series [pic] converges. Thus, the series

[pic] = [pic] converges.

Answer: Converges; Integral Test

3. [pic]

Since [pic] = [pic] = [pic] = [pic], then the series diverges by the Divergence Test.

Answer: Diverges; Divergence Test

4. [pic]

Let [pic]

The function f is positive on the closed interval [pic]. Now, we just need to check that f is decreasing on the interval [pic].

[pic] = [pic] = [pic]

[pic], which are not in the closed interval [pic]

[pic] is defined for all real numbers

Sign of [pic]: [pic]

(

1

Since [pic] for all x in the interval [pic], then the function f is decreasing on the closed interval [pic].

Thus, we can apply the Integral Test.

We need to evaluate the improper integral [pic].

[pic] = [pic]

The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.

[pic] = [pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic]

Thus,

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic]

Thus, the improper integral [pic] converges. Thus, by the Integral Test, the series [pic] converges.

Answer: Converges; Integral Test

5. [pic]

Let [pic]

[pic] for all [pic]

In order to show that the sequence [pic] is decreasing for all [pic], we need to show that [pic] for all [pic]. That is, we need to show that [pic] for all [pic].

Since [pic] for all [pic], then [pic] for all [pic] since [pic] is an increasing function of the open interval [pic].

Since [pic] is an increasing function of the closed interval [pic] and [pic] for all [pic], then [pic] for all [pic].

Since [pic] and [pic] for all [pic], then [pic] for all [pic].

Since [pic] and [pic] for all [pic], then [pic] for all [pic].

You can also establish the inequality that [pic] for all [pic] in the following manner. Since [pic] = [pic] and [pic] for all [pic], then [pic] = [pic]

Since [pic] for all [pic] and [pic] for all [pic], then [pic] for all [pic].

Since [pic] is a decreasing function of the open interval [pic] and [pic] for all [pic], then [pic] for all [pic].

Thus, [pic] for all [pic]. Thus, [pic] for all [pic]. Thus, the sequence [pic] is decreasing for all [pic].

Using Calculus to show that the sequence [pic] is decreasing for all [pic], we have the following.

Let [pic]. Then [pic]

[pic] =

[pic] = [pic] =

[pic] = [pic]

[pic], which is not in the closed interval [pic]

[pic] is defined when [pic]

[pic] or [pic]; [pic]. The numbers of 0 and 1 are not in the closed interval [pic].

Sign of [pic]: [pic]

(

2

Since [pic] for all x in the interval [pic], then the function f is decreasing on the closed interval [pic].

Thus, we can apply the Integral Test.

We need to evaluate the improper integral [pic].

[pic] = [pic]

The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.

[pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] = [pic]

Thus,

[pic] = [pic] = [pic] =

[pic] since [pic] since [pic]

Thus, the improper integral [pic] diverges. Thus, by the Integral Test, the series [pic] diverges.

Answer: Diverges; Integral Test

Theorem The p-series [pic] = [pic] converges if [pic] and diverges for [pic].

Proof Use the Integral Test. Will be provided later.

Example The series [pic] is a convergent p-series, where [pic].

Example The series [pic] is a divergent p-series, where [pic].

Theorem (The Basic Comparison Test) Suppose [pic] and [pic] are positive term series.

1. If the series [pic] converges and [pic] for all [pic], then the series [pic] converges.

NOTE: This statement is saying that if you can bound a series from above by a convergent series, then this series must also converge.

2. If the series [pic] diverges and [pic] for all [pic], then the series [pic] diverges.

NOTE: This statement is saying that if you can bound a series from below by a divergent series, then this series must also diverge.

Proof Will be provided later.

Examples Determine whether the following series converge or diverge using the Basic Comparison Test (BCT) if applicable.

1. [pic]

We established in an earlier example that this series converges by the Integral Test.

Since [pic] for all [pic], then [pic] for all [pic] since the function [pic] is a decreasing function on the open interval [pic].

Since [pic] for all [pic], then [pic].

Since the series [pic] is a convergent p-series with [pic], then the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

2. [pic]

Since [pic] for all [pic], then [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since the series [pic] is a convergent p-series with [pic], then the series

[pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

3. [pic]

Since [pic] for all [pic], then [pic] for all [pic] since the function [pic] is a decreasing function on the open interval [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is the divergent harmonic series, then the series [pic] is also divergent. Then the series [pic] = [pic] is also divergent.

Thus, the series [pic] diverges by the Basic Comparison Test (BCT).

Answer: Diverges; BCT

NOTE: We bounded the given series [pic] from below by a divergent series.

4. [pic]

Since [pic] for all positive odd integers and [pic] for all positive even integers, then we may write [pic] for all [pic]. Then [pic] for all [pic]. Thus, [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a divergent p-series with [pic], then the series [pic] = [pic] is divergent.

Thus, the series [pic] diverges by the Basic Comparison Test (BCT).

Answer: Diverges; BCT

NOTE: We bounded the given series [pic] from below by a divergent series.

5. [pic]

Since [pic], then [pic] = [pic]

Since [pic] for all positive odd integers and [pic] for all positive even integers, then we may write [pic] for all [pic]. Then [pic] for all [pic]. Thus, [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a convergent p-series with [pic], then the series [pic] = [pic] is convergent.

Thus, the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

6. [pic]

Since [pic] = [pic], then we can check the convergence or divergence of the series [pic]

Since [pic] for all [pic], then [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a convergent p-series with [pic], then the series [pic] = [pic] is convergent.

Thus, the series [pic] converges by the Basic Comparison Test (BCT).

Since the series [pic] is convergent, then the series [pic] = [pic] is convergent.

Answer: Converges; BCT

7. [pic]

Since [pic], then [pic] for all [pic]. Thus, [pic] for all [pic]. Thus, [pic] for all [pic]. Since [pic], then [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since the series [pic] is a convergent geometric series with [pic], then the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

8. [pic]

Since [pic] for all [pic], then [pic] for all [pic]. Thus, [pic] for all [pic]. Then [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a divergent p-series with [pic], then the series [pic] = [pic] is divergent.

Thus, the series [pic] diverges by the Basic Comparison Test (BCT).

Answer: Diverges; BCT

NOTE: We bounded the given series [pic] from below by a divergent series.

9. [pic]

Since [pic], then [pic] for all [pic] since the function [pic] is an increasing function on the open interval [pic]. Since the function [pic] is an increasing function for all real numbers and [pic] for all [pic], then [pic] for all [pic]. Since [pic], then [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is the divergent harmonic series and [pic], then the series [pic] is also divergent.

Thus, the series [pic] diverges by the Basic Comparison Test (BCT).

Answer: Diverges; BCT

NOTE: We bounded the given series [pic] from below by a divergent series.

10. [pic]

Since [pic], then [pic] for all [pic]. Thus, [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a convergent geometric series with [pic], then the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

11. [pic]

Since [pic] for all [pic], then [pic] for all [pic]. Thus, [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a convergent geometric series with [pic], then the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

12. [pic]

Since [pic] for all [pic], then [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic] is a divergent geometric series with [pic], then the series [pic] diverges by the Basic Comparison Test (BCT).

Answer: Diverges; BCT

NOTE: We bounded the given series [pic] from below by a divergent series.

13. [pic]

Since [pic] for all [pic], then [pic] for all [pic]. Since [pic] for all [pic], then [pic] for all [pic].

Since [pic] for all [pic], then [pic].

Since [pic], then [pic] = [pic]. The series [pic] is a convergent series since it is the sum of two convergent geometric series with [pic] and [pic]. Thus, the series [pic] is a convergent series. Since [pic], then the series [pic] converges by the Basic Comparison Test (BCT).

Answer: Converges; BCT

NOTE: We bounded the given series [pic] from above by a convergent series.

Theorem (The Limit Comparison Test) Suppose [pic] and [pic] are positive term series. If [pic], then either both series converge or both diverge.

Proof Will be provided later.

NOTE: When we apply the Limit Comparison Test, the series [pic] will be the series, which we want to know converges or diverges. The convergence or divergence of the series [pic], which we use for the comparison, is known. In most cases, the series [pic] will be a p-series.

Examples Determine whether the following series converge or diverge using the Limit Comparison Test (LCT) if applicable.

1. [pic]

We established that this series converges by the Integral Test and the Basic Comparison Test in earlier examples above.

Consider the series [pic] , which is a convergent p-series with [pic].

Let [pic] and [pic].

Then [pic] = [pic] = [pic] = [pic] = [pic] =

[pic].

Thus, the series [pic] converges by the Limit Comparison Test (LCT).

Answer: Converges; LCT

2. [pic]

We established that this series converges by the Basic Comparison Test in an earlier example above.

Consider the series [pic] , which is the divergent harmonic series.

Let [pic] and [pic].

Then [pic] = [pic] = [pic] = [pic] = [pic] =

[pic].

Thus, the series [pic] diverges by the Limit Comparison Test (LCT).

Answer: Diverges; LCT

3. [pic]

[pic] = [pic] = [pic] = [pic]

Thus, the series [pic] diverges by the Divergence Test (Div Test).

Answer: Diverges; Div Test

4. [pic]

Consider the series [pic] , which is a convergent p-series with [pic].

Let [pic] and [pic].

Then [pic] = [pic] = [pic] =

[pic] = [pic]

Thus, the series [pic] converges by the Limit Comparison Test (LCT).

Answer: Converges; LCT

5. [pic]

Consider the series [pic] , which is a divergent p-series with [pic].

Let [pic] and [pic].

Then [pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic] =

[pic]

Thus, the series [pic] diverges by the Limit Comparison Test (LCT).

Answer: Diverges; LCT

6. [pic]

Consider the series [pic] , which is a convergent p-series with [pic].

Let [pic] and [pic].

Then [pic] = [pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic]

Thus, the series [pic] converges by the Limit Comparison Test (LCT).

Answer: Converges; LCT

7. [pic]

Consider the series [pic] , which is the divergent harmonic series.

Let [pic] and [pic].

Then [pic] = [pic] = [pic] =

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

Thus, the series [pic] diverges by the Limit Comparison Test (LCT).

Answer: Diverges; LCT

8. [pic]

Consider the series [pic] , which is a convergent p-series with [pic].

Let [pic] and [pic].

Then [pic] = [pic]

Using L’Hopital’s Rule, we have that

[pic] = [pic] =

[pic] = [pic] = [pic] =

[pic]

Thus, [pic] = [pic]

Thus, the series [pic] converges by the Limit Comparison Test (LCT).

Answer: Converges; LCT

9. [pic]

[pic] = [pic] = [pic]

Thus, the series [pic] diverges by the Divergence Test (Div Test).

Answer: Diverges; Div Test

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