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I'm trying to find an explicit formula for the aforementioned sum (I am not sure that it is possible).

The first values are as follows, with $S_{k}=\sum_{n=0}^{\infty}\frac{n^k}{n!}$ :

$S_0=e$, $S_1=e$, $S_2=2e$, $S_3=5e$, $S_4=15e$, $S_5=52e$.

I couldn't see any obvious pattern from this, but I was able to find a recurrence relation for $S_k$ , by noticing that

$S_{k+2}=\sum_{n=1}^{\infty}\frac{n^{k}\left ( n-1+1 \right )}{\left ( n-1 \right )!}$ $=S_{k+1}+\sum_{n=1}^{\infty}\frac{\left ( n+1 \right )^{k}}{\left ( n-1 \right )!}$.

By expanding $(n+1)^{k}$, we get $S_{k+2}=S_{k+1}+\sum_{i=0}^{k}\binom{k}{i}S_{i+1}$ .

Problem is, I am unable to solve this recurrence relation.

It kind of reminds me of the relation between Bernoulli's numbers : $\sum _{k=0}^{n}{n+1 \choose {k}}B_{k}=0$. It is possible to find an explicit formula for Bernoulli's numbers, but it seems rather complicated to me (see this thread Explicit formula for Bernoulli numbers by using only the recurrence relation )

Perhaps it is possible to use a similar method to find an explicit formula for $S_k$ ?

EDIT : Thanks for the help. We can take the derivative inside the series since f is a power series with an infinite radius of convergence.

f is the exponential function so $f'=f$ :

$xf'\left ( x \right )=xf\left ( x \right )$,

$x\frac{\mathrm{d}}{\mathrm{d} x}\left ( xf'\left ( x \right ) \right )=\left ( x^{2}+x \right )f\left ( x \right )$,

$x\frac{\mathrm{d}}{\mathrm{d} x}\left (\left ( x^{2}+x \right )f\left ( x \right ) \right )=\left ( x^{3}+3x^{2}+x \right )f\left ( x \right )$

I can't seem to find a general formula for $\left ( x\frac{\mathrm{d} }{\mathrm{d} x} \right )^{k}f\left ( x \right )$ though. I'm pretty sure it involves binomial coeffficients...

If we have one, I see that by plugging $x=1$ we get a formula for $S_k$

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    $\begingroup$ You might find this helpful: en.wikipedia.org/wiki/Bell_number $\endgroup$ – carmichael561 Apr 12 '18 at 4:53
  • $\begingroup$ @charmichael561 thanks a lot. So $S_k=eB_k$... I'm sad to learn that we don't have any explicit formula for Bell numbers :( $\endgroup$ – Skywear Apr 12 '18 at 5:34
  • $\begingroup$ @AlexD Funny indeed ! That's just a question that I asked myself. I started calculating $\sum_{n=0}^{\infty}\frac{n}{n!}$, $\sum_{n=0}^{\infty}\frac{n^{2}}{n!}$ for no specific reason, and then wondered if I could find a more general formula. What about you ? $\endgroup$ – Skywear Apr 12 '18 at 5:38
  • $\begingroup$ en.wikipedia.org/wiki/Dobi%C5%84ski%27s_formula $\endgroup$ – Angina Seng Apr 12 '18 at 5:45
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    $\begingroup$ Also: en.wikipedia.org/wiki/Touchard_polynomials (evaluated at $x=1$) $\endgroup$ – Jack D'Aurizio Apr 12 '18 at 10:32
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Let $$f(x)=\sum_{n\geq 0} \frac{x^n}{n!}$$ And note that

$$xf'(x) =\sum_{n\geq 0} \frac{nx^n}{n!}$$

Repeat $k$ times and you'll have

$$(x\frac{d}{dx})^k f(x)=\sum_{n\geq 0} \frac{n^kx^n}{n!}.$$

Now, what is $f(x)$? And $f(1)$? Also, can you justify taking the derivatives inside the series?

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