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I've tried to understand why $\displaystyle\sum_{k=0}^{\infty} \frac{k^x}{k!}$ for lets say $x = 4$ equals $15e$.

It's clear why $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$ and that $\displaystyle\sum_{k=0}^{\infty} \frac{1^k}{k!}=e$

It's also unclear for me why $\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}=e$

I've tried to argue that $\displaystyle\sum_{k=0}^{\infty} \frac{e^k}{k!}= \displaystyle\sum_{k=0}^{\infty}\frac{k}{\ln(k!)}$ but that doesn't bring me further.

Hope someone here has got an idea for me

Thanks.

Boris Valderrama
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Quotenbanane
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    Maybe this question could help you understand how to prove the result https://math.stackexchange.com/questions/503451/infinite-series-sum-limits-k-1-infty-fracknk – Larry Jan 22 '19 at 23:59
  • Thank you! It's worth noting that i have searched in the search bar for this series, no clue why you found it and i didn't – Quotenbanane Jan 23 '19 at 00:01
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    It's quite a funny story. I have the same experience with you. I asked a question similar to yours, and people suggested me the link. I also didn't find this link when I tried to search it. Still, it is a good question. – Larry Jan 23 '19 at 00:05
  • Seems like a broken search system. Glad to have people like you! – Quotenbanane Jan 23 '19 at 13:33

1 Answers1

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It is very more simple than you think, it is only a recursive propertie. When $x=1$

$$ \sum_{k=0}^{\infty}\frac{k}{k!}=\sum_{k=1}^{\infty}\frac{k}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k}{k!}=\sum_{k=1}^{\infty}\frac{k}{(k-1)!\,k} $$ $$ \sum_{k=0}^{\infty}\frac{k}{k!}=\sum_{k=1}^{\infty}\frac{1}{(k-1)!} $$ $$ \sum_{k=0}^{\infty}\frac{k}{k!}=\sum_{k=0}^{\infty}\frac{1}{k!}=e $$ When $x=2$:

$$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{(k-1)!\,k} $$ $$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=\sum_{k=1}^{\infty}\frac{k}{(k-1)!} $$ $$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=\sum_{k=0}^{\infty}\frac{k+1}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=\sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} $$ From $x=1$: $$ \sum_{k=0}^{\infty}\frac{k^2}{k!}=e+e=2e $$ When $x=3$ $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=\sum_{k=1}^{\infty}\frac{k^3}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=\sum_{k=1}^{\infty}\frac{k^3}{(k-1)!\,k} $$ $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{(k-1)!} $$ $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=\sum_{k=0}^{\infty}\frac{(k+1)^2}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=\sum_{k=0}^{\infty}\frac{k^2}{k!}+2\sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} $$ From $x=2$ and $x=1$: $$ \sum_{k=0}^{\infty}\frac{k^3}{k!}=2e+2e+e=5e $$ When $x=4$: $$ \sum_{k=0}^{\infty}\frac{k^4}{k!}=\sum_{k=1}^{\infty}\frac{k^4}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^4}{k!}=\sum_{k=1}^{\infty}\frac{k^3}{(k-1)!} $$ $$ \sum_{k=0}^{\infty}\frac{k^4}{k!}=\sum_{k=0}^{\infty}\frac{(k+1)^3}{k!} $$ $$ \sum_{k=0}^{\infty}\frac{k^4}{k!}=\sum_{k=0}^{\infty}\frac{k^3}{k!}+3\sum_{k=0}^{\infty}\frac{k^2}{k!}+3\sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} $$ From $x=1,2,3$: $$ \sum_{k=0}^{\infty}\frac{k^4}{k!}=5e+6e+3e+e=15e $$ And that's all.

Boris Valderrama
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    My hero!!!! I also came up with sum{k=1}(1/(k-1)! but I'd not have thought of changing the indices. Again, thanks! – Quotenbanane Jan 23 '19 at 13:31