# Use the alternating series test to study the convergence of the following series sum_{n=1}^infty(-1)^{n+1}ne^{-n}

Use the alternating series test to study the convergence of the following series
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}n{e}^{-n}$
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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:
$|{S}_{k}-L|\le |{S}_{k}-{S}_{k+1}|={a}_{k+1}$
i) ${a}_{n}=n{e}^{-n}$
$=\frac{n}{{e}^{n}}>0$
ii) $n\to \mathrm{\infty }\frac{n}{{e}^{n}}$
$n\to \mathrm{\infty }\frac{1}{{e}^{n}}$
$=\frac{1}{\mathrm{\infty }}$
$=0$
iii)  ${a}_{n+1}$

Hence the alternating series will converge.

Jeffrey Jordon