2
$\begingroup$

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

$\endgroup$
  • 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
5
$\begingroup$

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

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

then

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

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.