4
$\begingroup$

Is there a formula for the closed form for $ \displaystyle \sum_{r=1}^\infty \frac{\sum_{k=1}^r k^n}{r!}$ for any positive integer $n$?

I tried Faulhaber's formula and Bell number but couldn't proceed.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

You can have a finite sum in terms of Bernoulli and Bell numbers. First we exploit Faulhaber's formula as

$$ \sum_{k=1}^r k^n =\frac{1}{n+1} \sum_{j=0}^n (-1)^j {n+1 \choose j} B_j\, r^{n+1-j} . $$

Then we have

$$ \sum_{r=1}^{\infty} \frac{1}{r!}\sum_{k=1}^r k^n = \frac{e}{n+1} \sum_{j=0}^n (-1)^j {n+1 \choose j} B_j Bell_{n-j+1} $$

$\endgroup$
10
  • $\begingroup$ @hypergeometric: It is a typo. $\endgroup$
    – science
    Apr 17, 2015 at 15:34
  • $\begingroup$ Why would this be a closed formula, as the OP asks for? $\endgroup$
    – Alex M.
    Apr 17, 2015 at 15:36
  • $\begingroup$ OK. In the second summation each term of $\frac 1{r!}$ is multiplied by the first equation which is a function of $r$, so can you isolate the two summations and replace the first one by $e$? $\endgroup$ Apr 17, 2015 at 15:37
  • $\begingroup$ @hypergeometric: When you change the order of summations you end up with the sum $$ \sum_{r=1}^{\infty}\frac{r^{n-i+1}}{r!} = Bell_{n-j+1}$$. $\endgroup$
    – science
    Apr 17, 2015 at 15:37
  • 1
    $\begingroup$ Yes, I know. Just clarifying for completeness for other readers here. $\endgroup$ Apr 17, 2015 at 16:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .