I would like to compute the following summation of series: $$F(k,a)=\sum_{m=-1}^{k}(-1)^m\frac{k!}{(k-m)!}a^{k-m},$$ where $a$ is a known constant, $k$ and $m$ are integers. $x!$ indicates the factorial value of $x$.

WolframAlpha gave me a result: $$F(k,a)=(-1)^ke^{-a}\Gamma(k+1,-a)-\frac{a^{k+1}}{k+1},$$ considering k is an integer: $$\Gamma(k+1,-a)=k!e^a\cdot e_k(-a),$$ then $$F(k,a)=(-1)^kk!e_k(-a)-\frac{a^{k+1}}{k+1},$$ here $e_k(x)$ is the exponentional sum function: $e_k(x)=\sum_{n=0}^{k}\frac{x^n}{n!}.$

I have checked that this result is correct for $k=0$. When assuming that it also holds for $k=N$, how to prove the correctness for $k=N+1$ using the constructive proof method?

Thanks very much indeed!

  • $\begingroup$ Did you try Wolframalpha? $\endgroup$ – InterstellarProbe Jun 22 '18 at 21:06
  • $\begingroup$ Why is the sum starting at -1? $\endgroup$ – herb steinberg Jun 22 '18 at 21:22
  • $\begingroup$ @InterstellarProbe Thanks for your notification! Although WolframAlpha could give a result, it doesn't post the inner steps and I couldn't prove its correctness. $\endgroup$ – yuhou CHEN Jun 22 '18 at 21:29
  • $\begingroup$ @herbsteinberg Well, it's just a part of my complicated computation and this summation should start at -1 for my problem. $\endgroup$ – yuhou CHEN Jun 22 '18 at 21:33
  • $\begingroup$ @InterstellarProbe I have re-edited the question, could you help me to prove the correctness of the WolframAlpha's result? Thanks a lot! $\endgroup$ – yuhou CHEN Jun 22 '18 at 22:14

Changing the summation index to $\,n=k-m\,$ gives the following, equivalent to the posted form:

$$ \begin{align} F(k,a) &= \sum_{n=0}^{k+1}\,(-1)^{k-n}\,\frac{k!}{n!}\,a^{n} \\[5px] &= (-1)^k\, k!\, \sum_{n=0}^{k+1}\,\frac{(-1)^{-n}a^{n}}{n!} \\[5px] &= (-1)^k\, k!\, \sum_{n=0}^{k+1}\, \frac{(-a)^{n}}{n!} \\[5px] &= (-1)^k \,k!\, e_{k+1}(-a) \end{align} $$


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