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.

Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$

I sense the answer has some connection with $e$, but I don't know how it is. Please help. Thank you.

share|improve this question
    
You can use Bell numbers. See here. –  Mhenni Benghorbal Oct 27 '12 at 8:25

5 Answers 5

up vote 6 down vote accepted

The most elementary calculation is probably this one:

$$\begin{align*} \sum_{k\ge 0}\frac{(k+1)^2}{k!}&=1+\sum_{k\ge 1}\frac{k^2+2k+1}{k!}\\ &=\color{red}{1}+\sum_{k\ge 1}\frac{k}{(k-1)!}+2\sum_{k\ge 1}\frac1{(k-1)!}+\color{red}{\sum_{k\ge 1}\frac1{k!}}\\ &=\sum_{k\ge 1}\frac{k-1+1}{(k-1)!}+2\sum_{k\ge 0}\frac1{k!}+\color{red}{\sum_{k\ge 0}\frac1{k!}}\\ &=\color{blue}{1}+\sum_{k\ge 2}\frac1{(k-2)!}+\color{blue}{\sum_{k\ge 2}\frac1{(k-1)!}}+3\sum_{k\ge 0}\frac1{k!}\\ &=\color{blue}{\sum_{k\ge 0}\frac1{k!}}+4\sum_{k\ge 0}\frac1{k!}\\ &=5\sum_{k\ge 0}\frac1{k!}\\ &=5e\;. \end{align*}$$

One can also make use of the identity $$x^n=\sum_k{n\brace k}x^{\underline k}\;,$$ where ${n\brace k}$ is a Stirling number of the second kind and $x^{\underline k}\triangleq x(x-1)(x-2)\dots(x-k+1)$ is a falling power. In particular,

$$\begin{align*} k^3&=\sum_{k=0}^3{3\brace i}k^i\\ &=0\cdot k^{\underline 0}+1\cdot k^{\underline 1}+3\cdot k^{\underline 2}+1\cdot k^{\underline 3}\\ &=k+3k(k-1)+k(k-1)(k-2)\;, \end{align*}$$

whence

$$\begin{align*} \frac{k^2}{(k-1)!}&=\frac{k^3}{k!}=\frac{k+3k(k-1)+k(k-1)(k-2)}{k!}\\ &=\frac1{(k-1)!}+\frac3{(k-2)!}+\frac1{(k-3)!}\;, \end{align*}$$

and summing over $k$ yields $5e$ as before.

share|improve this answer

Let's start with $e^x=\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!}$ then : $$x\,e^x=\sum_{k=1}^\infty \frac{x^k}{(k-1)!}$$ $$x(x\,e^x)'=x\,e^x+x^2\,e^x=\sum_{k=1}^\infty \frac{k\,x^k}{(k-1)!}$$ $$x(x(x\,e^x)')'=x(e^x+3x\,e^x+x^2\,e^x)=\sum_{k=1}^\infty \frac{k^2\,x^k}{(k-1)!}$$ Set $x=1$ to get : $$5\,e=\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$$

share|improve this answer
    
Very nice, this way seems the most clear to me. –  littleO Oct 27 '12 at 7:49
1  
Thanks @littleO ! The trick of using the theta operator $\theta=x\frac d{dx}$ works pretty often (two times here). –  Raymond Manzoni Oct 27 '12 at 7:52
    
Very Nice! Very Nice! Magic trick. –  Babak S. Oct 27 '12 at 9:16
    
Thanks @Babak generating functions tricks are indeed very nice! –  Raymond Manzoni Oct 27 '12 at 9:29
    
Excellent! Very insightful. –  Thomas Oct 27 '12 at 13:14

For $\frac{P(n)}{(n-r)!},$ where $P(n)$ is a polynomial.

If the degree of $P(n)$ is $m>0,$ we can write $P(n)=A_0+A_1(n-r)+A_2(n-r)(n-r-1)+\cdots+A_m(n-r)(n-r-1)\cdots\{(n-r)-(m-1)\}$

Here $k^2=C+B(k-1)+A(k-1)(k-2)$

Putting $k=1$ in the above identity, $C=1$

$k=2,B+C=4\implies B=3$

$k=0\implies 2A-B+C=0,2A=B-C=3-1\implies A=1$

or comparing the coefficients of $k^2,A=1$

So, $$\frac{k^2}{(k-1)!}=\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}=\frac1{(k-3)!}+\frac3{(k-2)!}+\frac1{(k-1)!}$$

$$\sum_{k=1}^\infty \frac{k^2}{(k-1)!}=\sum_{k=1}^\infty \left(\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}\right)$$ $$=\sum_{k=3}^\infty \frac1{(k-3)!}+\sum_{k=2}^\infty \frac3{(k-2)!}+\sum_{k=1}^\infty \frac1{(k-1)!}$$ as $\frac1 {(-r)!}=0$ for $r>0$

$=e+3e+e=5e$

share|improve this answer
    
How do you think to create the first line? –  ᴊ ᴀ s ᴏ ɴ Oct 27 '12 at 7:36
    
@jasoncube, this is the de facto way of solving $\frac{P(n)}{(n-r)!}$ where $P(n)$ is a polynomial of $n$. Here the highest power of $n$ is $2,$ so, there will be $2+1=3$ constants. –  lab bhattacharjee Oct 27 '12 at 7:41
    
Exactly the way I approached this. –  robjohn Oct 27 '12 at 7:43

I have provided a proof of the general case here. The proof provided shows

$$ \sum_{k=0}^{\infty} \frac{k^n}{k!} = T_n \cdot e $$

where $n$ is the $n^{th}$ Bell number. Yours is the case $n = 3$. Just for completeness, the formula above is called Dobinski's formula.

share|improve this answer

Let's finish it in one line

$$\sum_{k=1}^\infty \frac{(k-2)(k-1)+3(k-1)+1}{(k-1)!}=5e$$

Chris.

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.