Evaluate this series:

[yxgao]

Evaluate this series: Sum from k = 1 to k = infinity k^2/k!

The answer is 2e.

Thanks so much!!

[Hurkyl]

What have you tried so far?

[yxgao]

I have no idea how to do this problem. Does it require more knowledge than second year college math? I thought it just requires some knowledge about series, but I can't seem to get the answer. It's been over 3 years since I took calculus so I don't remember much about series expansions. I've tried comparing the series to several common expansions but I can't find the right form or find the right starting point. Thanks!

[arcnets]

yxgao,
here's a hint:

e = \sum_{k=0}^\infty \frac{1}{k!}

Try to express your sum in terms of this...

[Hurkyl]

Have you ever heard of a trick called "integrating with respect to 1"?


Anyways, one of the first things I'd do is simplify the summand; something cancels.

[krab]

Here's a hint: replace the index k by k+1. You'll be able to cancel k+1 in numerator and denominator. The rest is easy.

[yxgao]

How do you simplify (n+1)/n! summed from n = 0 to n = infinity?

[Hurkyl]

distribute!

[yxgao]

I don't understand how to proceed from this step. What do you do with the (n+1) in the numerator? What is there to distribute? Can someone post a direct method to the solution?
Thanks.

[Hurkyl]

\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}

[yxgao]

Ah, ok. Why didn't I see that before? :)
e + e = 2e
Thanks a lot.

[Hurkyl]

It's a slick trick, I haven't seen this particular sum done this way before. The method I know can be done as follows (it's more complicated, but more powerful):


The taylor series for e^x is:

e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}

I can insert a k term by differentiating:


e^x = \sum_{k=0}^{\infty} \frac{k x^{k-1}}{k!}


I can then multiply by x:

x e^x = \sum_{k=0}^{\infty} \frac{k x^{k}}{k!}

and differentiate again


(1 + x)e^x = \sum_{k=0}^{\infty} \frac{k^2 x^{k-1}}{k!}


Then, plug in x = 1 to get the sum.