3
$\begingroup$

By the ratio test the series $$ \sum_{n\ge0}\frac{n^x}{n!} $$ is convergent, but I know no method to evaluate it.

Since it's a convergent series then my question here is:

Is there a closed form for $\displaystyle\sum_{n\geq 0}\frac{n^x}{n!}$ with $x$ a positive real number?

$\endgroup$
0
$\begingroup$

I don’t think, that there is a closed form. The question should be changed to: "What interesting features does this series have (e.g. what integral representations are known) ?"

A note:

Be $\enspace\displaystyle f_{N,x}(z):= \sum\limits_{n=0}^N {\binom{N}{n}}n^x z^n\enspace $ with $\,x\geq 0\,$ , $\,z\in\mathbb{C}\,$ and $\,N\in\mathbb{N}_0\,$ .

It follows $\enspace\displaystyle \sum\limits_{n=0}^\infty \frac{n^x}{n!} = \lim\limits_{N\to\infty} f_{N,x}(\frac{1}{N}) \,$ .

For $\enspace x\in\mathbb{N}\enspace$ we have $\enspace\displaystyle f_{N,x}(z) = Nz(z+1)^{N-x}\sum\limits_{k=0}^{x-1}z^k \sum\limits_{v=0}^k {\binom{x-N}{k-v}}{\binom{N-1}{v}}(v+1)^{x-1}$

and we see that we get troubles generalizing (e.g) this to real $\,x> 0\,$ .

A hint $\,$ for $\,x\in\mathbb{N}\,$ :

$\displaystyle \sum\limits_{n=0}^\infty \frac{z^n n^x}{n!} = \lim\limits_{N\to\infty} f_{N,x}(\frac{z}{N}) =$

$\enspace\displaystyle = \lim\limits_{N\to\infty} (1+\frac{z}{N})^{N-x} z \lim\limits_{N\to\infty} \sum\limits_{k=0}^{x-1}(\frac{z}{N})^k \sum\limits_{v=0}^k {\binom{x-N}{k-v}}{\binom{N-1}{v}}(v+1)^{x-1}$

$\enspace\displaystyle =e^z \sum\limits_{k=0}^{x} S_{x,k} z^k\enspace$ with the Stirling numbers of the $2^{nd}$ kind $\,\displaystyle S_{x,k}:= \sum\limits_{v=0}^k \frac{(-1)^{k-v} v^x}{(k-v)!v!}\,$ .

The (first) series and the last sum of the equation chain are equal also for $\,x=0\,$ and $\,0^0:=1\,$.

$\endgroup$
  • $\begingroup$ Thanks for this answer because i don't got any help from above comments , i think this series will make a sense when x is a complex variable $\endgroup$ – zeraoulia rafik Jan 8 '18 at 16:48
  • $\begingroup$ @zeraouliarafik : It doesn't change anything, if you use $x$ as a complex variable. But the formula for $f_{N,x}(z)$ leads to the (non-trivial) formula for $\displaystyle\sum\limits_{n=0}^{\infty} \frac{n^x z^n}{n!}$ if $x\in\mathbb{N}$ . Because of this it's possible to see, which problem we have, if $x$ isn't a positive integer. $\endgroup$ – user90369 Jan 8 '18 at 22:01
  • $\begingroup$ you are right ! my question In Mo has been reached ,then if you want to share this question in MO i'm agree probably we will find who give us enough informuation for this non-trivial formula $\endgroup$ – zeraoulia rafik Jan 8 '18 at 22:05
  • $\begingroup$ @zeraouliarafik : As Jack D'Aurizio has mentioned, you can discuss the Touchard polynomials with real index, maybe you can construct an approximation. And: With my answer I just wanted to put a more elementary view on the series which helps to understand, why the existence of a closed form is very unlikely. $\endgroup$ – user90369 Jan 9 '18 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.