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I need to show that the series $\sum_{n=1}^{\infty}\frac{c^{-n}}{n!}$ is convergent.

I invoked the limit comparison with the series $\sum_{n=1}^{\infty}\frac{c^n}{n!}$ which is absolutely convergent (and hence convergent).

I got $\sum_{n=1}^{\infty}\frac{\frac{c^{-n}}{n!}}{\frac{c^n}{n!}}=\sum_{n=1}^{\infty}\frac{1}{n!n!}$.

I am not sure where to go from here. Would it be correct to write that $\frac{1}{n!n!}\rightarrow0$ ($n\rightarrow\infty$), and hence by the limit comparison test the series is convergent? Is there a better way to go about doing this?

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    $\begingroup$ That's not how comparison test works. If you knew that $\sum c^n/n!$ converges absolutely, then $\sum (c^{-1})^n/n!$ converges as well. $\endgroup$ – xbh Jan 20 '19 at 10:38
  • $\begingroup$ A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $ $\endgroup$ – Milan Jan 20 '19 at 11:45
  • $\begingroup$ $$\frac{\frac{c^{-n}}{n!}}{\frac{c^n}{n!}}\ne\frac1{n!n!}$$ $\endgroup$ – Did Jan 31 '19 at 13:50
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Use the ratio test:$$\lim_{n\in\mathbb N}\frac{\frac{c^{-n-1}}{(n+1)!}}{\frac{c^{-n}}{n!}}=\lim_{n\in\mathbb N}\frac1{c\times(n+1)}=0<1.$$

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    $\begingroup$ Provided $c\ne0$. $\endgroup$ – Yves Daoust Jan 20 '19 at 10:46
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    $\begingroup$ If $c=0$, then the statement itself doesn't make sense. $\endgroup$ – José Carlos Santos Jan 20 '19 at 10:49
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Easier to notice that

$$ \sum \frac{ c^{-n} }{n!} = e^{1/c} $$

So as long as $c \neq 0$, we have convergence.

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    $\begingroup$ This is wrong. This series sums to $\exp(1/c)$, not $\exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$. $\endgroup$ – Clayton Jan 20 '19 at 10:59

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