# 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 $n$?

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.

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

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

• @hypergeometric: It is a typo. Apr 17, 2015 at 15:34
• Why would this be a closed formula, as the OP asks for? Apr 17, 2015 at 15:36
• 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 $e$? Apr 17, 2015 at 15:37
• @hypergeometric: When you change the order of summations you end up with the sum $$\sum_{r=1}^{\infty}\frac{r^{n-i+1}}{r!} = Bell_{n-j+1}$$. Apr 17, 2015 at 15:37
• Yes, I know. Just clarifying for completeness for other readers here. Apr 17, 2015 at 16:44