I have to find the sum of the following series: $$\sum\limits_{n=1}^\infty \frac{n^3}{n!}$$

I know how to prove the convergence of this series, but how do i find the sum. I can't use the properties of derivation or integration of the series, i have to do it using the definition that a series converges iff it's sequence of partial sum converges.


Expanding on @Winther's comment, you may write

$$ \begin{align} \sum_1^{\infty} \frac{n^3}{n!}&=\sum_1^{\infty} \frac{n (n - 1) (n - 2) + 3 n (n - 1) + n}{n!}\\\\ &=\sum_3^{\infty} \frac{n (n - 1) (n - 2)}{n!}+3\sum_2^{\infty} \frac{n(n - 1) }{n!}+\sum_1^{\infty} \frac{n}{n!}\\\\ &=\sum_3^{\infty} \frac{1}{(n-3)!}+3\sum_2^{\infty} \frac{1}{(n-2)!}+\sum_1^{\infty} \frac{1}{(n-1)!}\\\\ &=\sum_0^{\infty} \frac{1}{n!}+3\sum_0^{\infty} \frac{1}{n!}+\sum_0^{\infty} \frac{1}{n!}\\\\ &=5e \end{align} $$

since $$ \sum_0^{\infty} \frac{x^n}{n!}=e^x, \quad x \in \mathbb{C}. $$


In general, if

$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$


$$\sum_{n=1}^{\infty} n^3 a_n z^n = \frac{d}{dz} \left [z \frac{d}{dz} \left [z \frac{d}{dz} f(z) \right ]\right ]$$

In your case $a_n = 1/n!$ implies that $f(z) = e^z$. Take the derivatives, set $z=1$. The result is

$$\sum_{n=0}^{\infty} \frac{n^3}{n!} = 5 e$$

  • $\begingroup$ How can one find $f$ forn $a_n$? Just curious, if it is long, I'll ask a question about it. $\endgroup$ – wythagoras Oct 26 '15 at 17:22
  • $\begingroup$ @wythagoras: Well, it helps to be familiar with some basic Taylor series, such as that for $e^z$. $\endgroup$ – Ron Gordon Oct 26 '15 at 17:23
  • $\begingroup$ Thank you for the tip, but as I said I can't use derivatives or integrals, because in this curse I'm learning only the basics (Analysis 1), in Analysis 2 I learn derivatives and integrals, so that is the main problem... $\endgroup$ – zermelovac Oct 26 '15 at 17:24
  • $\begingroup$ Then you may want to read the comment from @Winther. Again, that assumes you know how to recognize a famous series for $e$. $\endgroup$ – Ron Gordon Oct 26 '15 at 17:26

The other answers have given you different ways to solve the problem at hand. I will below just add some notes on how one can evaluate general sums on the form $\sum_{n=0}^\infty \frac{n^k}{n!}$ where $k$ is any integer using just algebraic manipulations (since you said you cannot use derivatives). I'm however assuming we already know that $\sum_{n=0}^\infty\frac{1}{n!} = e$.

We start by noting that we can always write $n^k$ on the form

$$n^k = C_1n + C_2n(n-1) + \ldots + C_kn(n-1)\cdots(n-k+1) \equiv \sum_{j=1}^{k} C_j{n\choose j}j!$$

where $C_i$ are some real numbers (see below for a simple algorithm to compute these numbers). For example

$$n^2 = n(n-1) + n$$ and $$n^4 = n(n-1)(n-2)(n-3) + 6 n (n - 1) (n - 2) + 7 n (n - 1) + n$$

The reason this form is useful is that we can very easily evaluate the sums

$$\sum_{n=0}^\infty\frac{n(n-1)\cdots(n-i)}{n!} = \sum_{n=i+1}^\infty \frac{1}{(n-i-1)!} = \sum_{m=0}^\infty \frac{1}{m!} = e$$

where we have simply shifted the summation index to $m = n-i-1$ in the last step. The sum above does not depend on $i$ at all so the sum we are after becomes

$$\sum_{n=0}^\infty \frac{n^k}{n!} = \sum_{n=0}^\infty \frac{ C_1n + C_2n(n-1) + \ldots + C_kn(n-1)\cdots(n-k+1)}{n!} \\= C_1 e + C_2 e + \ldots + C_k e = \left(\sum_{j=1}^k C_j\right) e$$

so the problem reduces to finding the $C_i$'s.

Here is a simple method to compute the $C_i$'s. Lets take $k=3$ as an example and start by writing $$n^3 = \color{blue}{1}n^3$$ Now we subtract $\color{blue}{1}n(n-1)(n-2)$ to get $$n^3 - \color{blue}{1} n(n-1)(n-2) = \color{red}{3} n^2 - 2n$$ so $C_2 = \color{red}{3}$. Now subtract $\color{red}{3}n(n-1)$ to get $$n^3 - n(n-1)(n-2) - \color{red}{3}n(n-1) = \color{pink}{1} n$$ so $C_1 = \color{pink}{1}$. We have now shown that $$n^3 = \color{blue}{1}n(n-1)(n-2) + \color{red}{3}n(n-1) + \color{pink}{1} n$$ so using the formula above we have $$\sum_{n=0}^\infty \frac{n^3}{n!} = (\color{blue}{1}+\color{red}{3}+\color{pink}{1})e = 5e$$

If you want to read more about this, the numbers $C_j$ are the so-called Stirling numbers of the second kind, $C_j \equiv S(k,j)$, and the sum $\sum_{j=1}^k C_j = \sum_{j=1}^k S(k,j) \equiv B_k$ are the so-called $k$'th Bell number (and is always an integer). Using this the final result can be compactly written

$$\sum_{n=0}^\infty \frac{n^k}{n!} = eB_k$$

which is known as Dobinski's formula. Knowing this there exist an even simpler method to compute the Bell numbers using a "Pascal's triangle" approach, see for example this page.


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