# How can I sum the series $e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$

How can I sum the following series?

$$e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$$

I think I can make this sum in the form of exponential expansion but not able to think how. Any initial hint would be great. Thanks in advance.

• that does not makes sense to start the sum at $x=0$ – Guy Fsone Jan 16 '18 at 13:10
• Are $x$ and $y$ both integers? If so, what is $(x-y)!$ for $y>x$? – 5xum Jan 16 '18 at 13:11
• @GuyFsone if $y> 0$ then $\frac1{(-y)!}=0$ because $|(-y)!|=\infty$ – Masacroso Jan 16 '18 at 13:16
• @Masacroso Why is $|(-y)!|=\infty$, what are you using to define the negative factorial? See also this. – Michael Burr Jan 16 '18 at 13:23
• @Masacroso how do you know that ? – Guy Fsone Jan 16 '18 at 13:23

$$\sum_{n=k}^{\infty}\frac{x^n}{(n-k)!}=x^k\sum_{n=k}^{\infty}\frac{x^{n-k}}{(n-k)!}=x^k\sum_{n=0}^{\infty}\frac{x^{n}}{n!} = x^ke^x$$
now take $x=\frac12$