# How do you do the Alternating Series Test on this series and what is the result? sum_{n=2}^inftyfrac{(-1)^n}{n+1}

How do you do the Alternating Series Test on this series and what is the result?
$\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{n+1}$
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Alternating series:
A series of the form
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}n{a}_{n}={a}_{0}-{a}_{1}+{a}_{2}-{a}_{3}+...$
where either all ${a}_{n}$ are positive or all ${a}_{n}$ are negative, is called an alternating series.
The alternating series test then says: if $|{a}_{n}|$ decreases monotonically and $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$ then the alternating series converges.
Moreover, let L denote the sum of the series, then the partial sum
${S}_{k}=\sum _{n=0}^{k}\left(-1{\right)}^{n}{a}_{n}$
approximates L with error bounded by the next omitted term:
$|{S}_{k}-L|\le |{S}_{k}-{S}_{k+1}|={a}_{k+1}$
${a}_{n}=\frac{1}{n+1}$
${a}_{n+1}=\frac{1}{n+2}$

Therefore ${a}_{n+1}<{a}_{n}$

So ${a}_{n}$ is monotonically decrea $\mathrm{sin}g$

${a}_{n}>0$ for all $n=1,2,3,...$

$n\to \mathrm{\infty }$

$=0$

So the series converges.

Jeffrey Jordon