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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.

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  • $\begingroup$ that does not makes sense to start the sum at $x=0$ $\endgroup$
    – Guy Fsone
    Commented Jan 16, 2018 at 13:10
  • $\begingroup$ Are $x$ and $y$ both integers? If so, what is $(x-y)!$ for $y>x$? $\endgroup$
    – 5xum
    Commented Jan 16, 2018 at 13:11
  • $\begingroup$ @GuyFsone if $y> 0$ then $\frac1{(-y)!}=0$ because $|(-y)!|=\infty$ $\endgroup$
    – Masacroso
    Commented Jan 16, 2018 at 13:16
  • $\begingroup$ @Masacroso Why is $|(-y)!|=\infty$, what are you using to define the negative factorial? See also this. $\endgroup$ Commented Jan 16, 2018 at 13:23
  • $\begingroup$ @Masacroso how do you know that ? $\endgroup$
    – Guy Fsone
    Commented Jan 16, 2018 at 13:23

1 Answer 1

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Basically

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

now take $x=\frac12$

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