Check Convergence of $\sum _{n+1}^{\mathrm{\infty}}(-1{)}^{n}\ast (\frac{{e}^{n}}{n!})$

Jaslyn Sloan
2022-11-19
Answered

Check Convergence of $\sum _{n+1}^{\mathrm{\infty}}(-1{)}^{n}\ast (\frac{{e}^{n}}{n!})$

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mainzollbtt

Answered 2022-11-20
Author has **13** answers

Let us put

${a}_{n}:=\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{{a}_{n+1}}{{a}_{n}}=\frac{{e}^{n+1}}{(n+1)!}\frac{n!}{{e}^{n}}=\frac{e}{n+1}\underset{n\to \mathrm{\infty}}{\overset{}{\to}}0$

Thus, the series converges absolutely and thus it also converges, i.e.:

$\sum _{n=1}^{\mathrm{\infty}}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{converges}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{\mathrm{\infty}}(-1{)}^{n}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{also converges}$

${a}_{n}:=\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{{a}_{n+1}}{{a}_{n}}=\frac{{e}^{n+1}}{(n+1)!}\frac{n!}{{e}^{n}}=\frac{e}{n+1}\underset{n\to \mathrm{\infty}}{\overset{}{\to}}0$

Thus, the series converges absolutely and thus it also converges, i.e.:

$\sum _{n=1}^{\mathrm{\infty}}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{converges}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{\mathrm{\infty}}(-1{)}^{n}\frac{{e}^{n}}{n!}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{also converges}$

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