# Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series

## Key Questions

• In most cases, an alternation series ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$ fails Alternating Series Test by violating ${\lim}_{n \to \infty} {b}_{n} = 0$. If that is the case, you may conclude that the series diverges by Divergence (Nth Term) Test.

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• Alternating Series Test

An alternating series ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, ${b}_{n} \ge 0$ converges if both of the following conditions hold.

$\left\{\begin{matrix}{b}_{n} \ge {b}_{n + 1} \text{ for all } n \ge N \\ {\lim}_{n \to \infty} {b}_{n} = 0\end{matrix}\right.$

Let us look at the posted alternating series.

In this series, ${b}_{n} = \frac{1}{\sqrt{3 n + 1}}$.

${b}_{n} = \frac{1}{\sqrt{3 n + 1}} \ge \frac{1}{\sqrt{3 \left(n + 1\right) + 1}} = {b}_{n + 1}$ for all $n \ge 1$.

and

${\lim}_{n \to \infty} {b}_{n} = {\lim}_{n \to \infty} \frac{1}{\sqrt{3 n + 1}} = \frac{1}{\infty} = 0$

Hence, we conclude that the series converges by Alternating Series Test.

I hope that this was helpful.

• Alternating Series Test states that an alternating series of the form
${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, where ${b}_{n} \ge 0$,
converges if the following two conditions are satisfied:
1. ${b}_{n} \ge {b}_{n + 1}$ for all $n \ge N$, where $N$ is some natural number.
2. ${\lim}_{n \to \infty} {b}_{n} = 0$

Let us look at the alternating harmonic series ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} \frac{1}{n}$.
In this series, ${b}_{n} = \frac{1}{n}$. Let us check the two conditions.
1. $\frac{1}{n} \ge \frac{1}{n + 1}$ for all $n \ge 1$
2. ${\lim}_{n \to \infty} \frac{1}{n} = 0$

Hence, we conclude that the alternating harmonic series converges.