# What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?

For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:

What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$

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The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2\exp(1)$.

In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.

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A basic technique in real (complex) analysis is term by term differentiation of power series:

$$e^z=\sum_{k=0}^\infty\frac{z^k}{k!},\quad e^z=(e^z)'=\sum_{k=1}^\infty k\cdot\frac{z^{k-1}}{k!},\quad e^z=(e^z)''=\sum_{k=1}^\infty k(k-1)\frac{z^{k-2}}{k!}.$$ Evaluating at $z=1$, one immediately has $$e=\sum_{k=0}^\infty\frac{1}{k!},\quad e=\sum_{k=1}^\infty k\cdot\frac{1}{k!},\quad e=\sum_{k=1}^\infty (k^2-k)\frac{1}{k!}.$$ Combining the second and third equalities, we have the answer.

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Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=\sum \limits_{n=0}^\infty \frac{x^n}{n!}$, apply $\frac{d}{dx}x\frac{d}{dx}$ to both sides, and evaluate at $x=1$.

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$\frac{d}{dx}x\frac{d}{dx}$, typo? – Jack Jun 8 '11 at 17:46
It means: differentiate, then multiply by $x$, then differentiate. – GEdgar Jun 8 '11 at 17:54

The value of $T_n := \displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}$ is $B_n \cdot e$, where $B_n$ is the $n^{th}$ Bell number.

To see this, note that

\begin{align} T_{n+1} = \sum_{k=1}^{\infty} \frac{k^{n+1}}{k!} &= \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^n {n \choose j} k^j \\ &= \sum_{j=0}^n {n \choose j} \sum_{k=1}^{\infty} \frac{k^j}{k!} \\ &= \sum_{j=0}^{n} {n \choose j} T_j \end{align}

This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.

Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.

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+1,I like this answer. – Eric Naslund Jun 8 '11 at 16:28
+1, nice for providing the background. – Jack Jun 8 '11 at 16:41
@Eric/Jack Thanks! – JavaMan Jun 8 '11 at 16:49

Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $x\mapsto\exp(x)$.

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