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I was wondering if there was a technique to find the infinite sum of the series below where n is a positive integer.

$\sum\limits_{k=n+1}^\infty (k-n)\frac{8.676^ke^{-8.676}}{k!} $

Thank you!

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    I only know of a way to calculate it for 8.677, sorry. – 2017-02-12
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    @mathworker21: Can you show your method for 8.677? – 2017-02-13

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$$\sum\limits_{k=n+1}^\infty (k-n)\frac{a^ke^{-a}}{k!}$$ $$\sum\limits_{k=1}^\infty k\frac{a^{n+k}e^{-a}}{(n+k)!}$$ $$a^ne^{-a}\sum\limits_{k=1}^\infty k\frac{a^k}{(n+k)!}$$ This sum isn't super easy to simplify; we have to use special functions. We get our final result as $$\frac{a^{n+1}(n-a) \operatorname{E}_{-n}(a)+a^{n+1}e^{-a}}{\Gamma (n+1)}+a-n$$ Where $\operatorname{E}_{-n}(.)$ is the exponential integral and $\Gamma(.)$ is the Gamma Function. Note that if your sum went from $0$ to $\infty$ we would instead just get $a-n$, and everything else above is the correction factor for your change of index

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    I understand how you got to the new sum, is there any chance you could explain a couple of the steps to get to the final result? – 2017-02-12