Absolute and Conditional Convergence: Definition, Examples

A series ∑a_n is called absolutely convergent if ∑|a_n| converges. A series ∑a_n is called conditionally convergent if it converges but not absolutely. In this article, we study absolute and conditional convergence with their definitions and examples.

Absolutely Convergent Series

Definition: A series $\displaystyle \sum_{n=1}^\infty a_n$ is called absolutely convergent if the series $\displaystyle \sum_{n=1}^\infty |a_n|$ is convergent.

For example, the series $1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+\cdots$ is an absolutely convergent series.

REMARK:

An absolutely convergent series is convergent. That is, if a series converges absolutely then it also converges. But, the converse is not true, that’s where the concept of conditionally convergent series comes into the picture.

Conditionally Convergent Series

Definition: A series $\displaystyle \sum_{n=1}^\infty a_n$ is called conditionally convergent (or semi-convergent) if the series is convergent but not absolutely.

For example, the series $1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots$ is a conditionally convergent series.

Questions and Answers

Q1: Show that the series $1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+\cdots$ is absolutely convergent.

Answer:

The series ∑a_n = $\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \dfrac{1}{n^2}$.

Now, ∑|a_n| = $\displaystyle \sum_{n=1}^\infty |(-1)^{n+1} \dfrac{1}{n^2}|$ = $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}$. This is a p-series with p=2>1; hence ∑|a_n| is convergent.

Therefore, the given series is absolutely convergent.

Q2: Show that the series $1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots$ is conditionally convergent.

Answer:

Step 1: (To show the series converges)

The series = ∑ (-1)ⁿ⁺¹a_n = $\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \dfrac{1}{2n-1}$.

Thus, a_n = $\dfrac{1}{2n-1}$. Now, we have:

1. $\dfrac{1}{2n-1} >0$ for all n=1,2,3,… That is, an > 0 ∀ n.

2. an – an+1 = $\dfrac{1}{2n-1} – \dfrac{1}{2(n+1)-1}$ = $\dfrac{2n+1-2n+1}{(2n-1)(2n+1)}$ = $\dfrac{2}{(2n-1)(2n+1)}$ > 0 ∀ n. So, an – an+1 > 0 ∀ n. This shows that {an} is monotonically decreasing.

3. limn→∞ an = limn→∞ $\dfrac{1}{2n-1}$ = 0.

So by Leibnitz’s test, the given series converges.

Step 2: (To show the series is not absolutely convergent)

Now, we will show that this series does not converge absolutely.

The series of absolute terms is given by $1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\cdots$ = $\displaystyle \sum_{n=1}^\infty x_n$, where x_n = $\dfrac{1}{2n-1}$.

We use the limit comparison test with the series ∑ y_n = ∑ 1/n. Now,

limn→∞ (xn/yn) = limn→∞ $\dfrac{n}{2n-1}$ = limn→∞ $\dfrac{1}{2-\frac{1}{n}}$ (dividing top and bottom by n) = 1/2, which is finite and positive.

Thus, by limit form of comparison test, both ∑ 1/n and ∑ 1/(2n-1) behave similarly. We know ∑ 1/n diverges by p-series as p=1. Hence, the series ∑ 1/(2n-1) also diverges.

Therefore, the given series is absolutely convergent and hence by definition conditionally convergent.

Q3: Show that the series $1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots$ is conditionally convergent.

Answer:

Step 1: (To show the series is not absolutely convergent)

The series = ∑ a_n = $\displaystyle \sum_{n=1}^\infty (-1)^{n+1} \dfrac{1}{n}$.

Note that ∑ |a_n| = $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}$ is not convergent by p-series as p=1.

So the series is not absolutely convergent.

Step 2: (To show the series converges)

The series converges by Leibnitz’s test. The proof is here.

Therefore, by definition, the given series is conditionally convergent.

Also Study:

• Convergence of a Series: Definition, Formula, Examples
• Comparison Test for Series: Statement, Examples [Limit Form]
• Ratio Test for Series: Statement, Solved Examples
• Root Test
• Alternating Series and Leibnitz’s Test

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