Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$

share|improve this question
    

4 Answers 4

up vote 35 down vote accepted

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 $e$.

share|improve this answer

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 $x=1$.

share|improve this answer
    
$\frac{d}{dx}x\frac{d}{dx}$, typo? –  Jack Jun 8 '11 at 17:46
9  
It means: differentiate, then multiply by $x$, then differentiate. –  GEdgar Jun 8 '11 at 17:54
    
I would be curious to know why the downvote. –  Ross Millikan Jul 7 '11 at 12:33

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.

share|improve this answer
3  
+1,I like this answer. –  Eric Naslund Jun 8 '11 at 16:28
3  
+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 $x\mapsto\exp(x)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.