Find the function of integer numbers

$$\sum_{n=0}^{\infty }\frac{n^k}{n!}={f(k)}\cdot e$$

I took many values of $k$ and I found the following results $$\sum_{n=0}^{\infty }\frac{n^1}{n!}=e$$ $$\sum_{n=0}^{\infty }\frac{n^2}{n!}=2e$$ $$\sum_{n=0}^{\infty }\frac{n^3}{n!}=5e$$ $$\sum_{n=0}^{\infty }\frac{n^4}{n!}=15e$$ $$\sum_{n=0}^{\infty }\frac{n^5}{n!}=52e$$ and so on

I think these numerical values are right, so I tried to find the function $f(k)$.

Can help me to find the function $f(k)$ and then prove the above series

  • 4
    $\begingroup$ A bit wrong as you state it, but this is Dobinski's formula. The integers are Bell numbers. The proof on Wikipedia, using factorial moments of Poisson distribution, is a classic. $\endgroup$ – Stop hurting Monica Dec 10 '14 at 20:31
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    $\begingroup$ Sorry but there seems to be a mistake. We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$. Hence, $\sum_{n=0}^{\infty} \frac{n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = 1 + \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + e^1$. $\endgroup$ – GenericNickname Dec 10 '14 at 20:32
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    $\begingroup$ You probably mean $e, 2e, 5e, \dots$ above, not $e, e^2, e^5, \dots$. $\endgroup$ – anomaly Dec 10 '14 at 20:33
  • $\begingroup$ yes yes yes ,I am sorry $\endgroup$ – E.H.E Dec 10 '14 at 20:34

Define the following sequence

$$h_{k+1}(x) = xh_k'(x)$$

with $h_0(x) = e^x$. Then (by induction)

$$h_k(x) = \sum_{n=0}^\infty \frac{n^kx^{n}}{n!}$$

and from this it follows (again by induction) that $$h_k(1) \equiv \sum_{n=0}^\infty \frac{n^k}{n!} = f(k) e$$ where $f(k)$ are integers. To find an expression for $f(k)$ define

$$g(x,t) = \sum_{n=0}^\infty \frac{h_{n}(x)t^n}{n!}$$


$$g(x,t) = \sum_{n=0}^\infty\sum_{k=0}^\infty \frac{(kt)^n}{n!k!}x^k = e^{xe^t}$$

and by taking $x=1$ it follows that

$$\sum_{n=0}^\infty \frac{f(n)x^n}{n!} = e^{e^x -1}$$

which is the exponential generation function for the Bell numbers.


The function $F(x) = e^x$ is given by $$F(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$ and thus has $k$th derivative $$F^{(k)}(x) = \sum_{n=0}^\infty n(n - 1) \cdots (n - (k-1)) \frac{x^n}{n!} = k!\sum_{n=0}^\infty \binom{n}{k} \frac{x^n}{n!}.$$ (I'm being very cavalier here about the sum, but standard uniform convergence arguments will make the approach rigorous.) But $F^{(k)}(x) = F(x)$, so $$\sum_{n=0}^\infty \binom{n}{k} \frac{x^n}{n!} = \frac{1}{k!} f^{(k)}(1) = \frac{e}{k!}.$$ Rewriting the polynomials $n^k$ in terms of the $\binom{n}{m}$ gives the required result.

  • $\begingroup$ And... you call the exponential $f$ because the question already uses $f(k)$ with a precise (different) meaning? $\endgroup$ – Did Dec 11 '14 at 8:45
  • $\begingroup$ Call it $F$ if you prefer. $\endgroup$ – anomaly Dec 11 '14 at 14:30
  • $\begingroup$ ...or, rather, I'll change it myself. :) $\endgroup$ – anomaly Dec 11 '14 at 18:46

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