Alternating Series Test

Tags

Calculus

Cegep/2

Word count

197 words

Reading time

2 minutes

Test to determine whether a series that alternates between positive and negative diverges

Let $\sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n\text{ or }\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n$.
Then if

1. $\underset{n\to \mathrm{\infty }}{lim}{b}_n=0$
2. $\{b_n{\}}_{n=1}^{\mathrm{\infty }}$ is decreasing for $n\in (N,\mathrm{\infty })$

then $\sum a_n$ is convergent.

Examples

Determine if the following series are convergent:

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n\cos (n\pi )}{n^3+1}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}& =\frac{(-1{)}^n\cos (n\pi )}{n^3+1}\\ & =\frac{(-1{)}^n(-1{)}^n}{n^3+1}\\ & =\frac{1}{n^3+1}\end{array}$$

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-5{)}^n}{(n+1)5^n}$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{(-5{)}^n}{(n+1)5^n}=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n+1}$$

Let $b_n=\frac{1}{n+1}$.

Since $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n+1}=0$ and $f(x=\frac{1}{x+1})⟹f^{\prime }(x)=-\frac{1}{(x+1{)}^2}<0\mathrm{\forall }x\in \mathbb{R}$,
By AST, ${b_n}_{n=1}^{\mathrm{\infty }}$ is decreasing for $n\ge 1$.

Contributors

Ajitani Hifumi

Changelog

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