# Proving that the series $\sum_{n=1}^{\infty}\frac{c^{-n}}{n!}$ is convergent

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?

• 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. – xbh Jan 20 '19 at 10:38
• A factorial is a good sign that ratio test can be used . $(n+1)!=n!(n+1)$ – Milan Jan 20 '19 at 11:45
• $$\frac{\frac{c^{-n}}{n!}}{\frac{c^n}{n!}}\ne\frac1{n!n!}$$ – Did Jan 31 '19 at 13:50

Use the ratio test:$$\lim_{n\in\mathbb N}\frac{\frac{c^{-n-1}}{(n+1)!}}{\frac{c^{-n}}{n!}}=\lim_{n\in\mathbb N}\frac1{c\times(n+1)}=0<1.$$
• Provided $c\ne0$. – Yves Daoust Jan 20 '19 at 10:46
• If $c=0$, then the statement itself doesn't make sense. – José Carlos Santos Jan 20 '19 at 10:49
$$\sum \frac{ c^{-n} }{n!} = e^{1/c}$$
So as long as $$c \neq 0$$, we have convergence.
• 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 $e^{1/c}-1$. – Clayton Jan 20 '19 at 10:59