# Check Convergence of sum_(n+1)^infty (-1)^n * (e^n/n!)

Jaslyn Sloan 2022-11-19 Answered
Check Convergence of $\sum _{n+1}^{\mathrm{\infty }}\left(-1{\right)}^{n}\ast \left(\frac{{e}^{n}}{n!}\right)$
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## Answers (1)

mainzollbtt
Answered 2022-11-20 Author has 13 answers
Let us put
${a}_{n}:=\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{{a}_{n+1}}{{a}_{n}}=\frac{{e}^{n+1}}{\left(n+1\right)!}\frac{n!}{{e}^{n}}=\frac{e}{n+1}\underset{n\to \mathrm{\infty }}{\overset{}{\to }}0$
Thus, the series converges absolutely and thus it also converges, i.e.:
$\sum _{n=1}^{\mathrm{\infty }}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{converges}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{also converges}$
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