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.
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.