Alternating Series, Leibnitz’s Test: Statement, Solved Examples

An alternating series is a series containing terms alternatively positive and negative. One can check its convergence using Leibnitz’s test. In this article, we will study alternating series, Leibnitz’s test with solved problems.

Definition of Alternating Series

A series of the form $\displaystyle \sum_{n=1}^\infty$ (-1)ⁿ⁺¹a_n, where a_n>0 for all n, is called an alternating series.

The series 1- 1/2+ 1/3- 1/4- … is an example of alternating series.

Statement of Leibnitz’s Test

Let {a_n} be a sequence of real numbers such that

1. a_n > 0 for all n.
2. {a_n} is monotonically decreasing. That is, a_n > a_n+1 ∀ n.
3. lim_n→∞ a_n =0.

Then the corresponding alternating series $\displaystyle \sum_{n=1}^\infty$(-1)ⁿ⁺¹a_n is convergent.

Questions and Answers

Q1: Test the convergence of $1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots$.

Answer:

The terms of the given series are alternatively positive and negative, so it is an alternating series.

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

Thus, a_n = $\dfrac{1}{n}$.

1. $\dfrac{1}{n} >0$ for all n. That is, an > 0 ∀ n.

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

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

Therefore, all the conditions of Leibnitz’s test are satisfied. Hence by Leibnitz’s test we conclude that the series 1-1/2+1/3-1/4-… is convergent.

Related Articles:

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

Homework:

Test the convergence of the following series:

1. $1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+\cdots$
2. $1-\dfrac{1}{2^p}+\dfrac{1}{3^p}-\dfrac{1}{4^p}+\cdots$ (p>0)
3. $\dfrac{1}{1+x^2}-\dfrac{1}{2+x^2}+\dfrac{1}{3+x^2}+\cdots$.

FAQs