# Evaluate this series:

1. Dec 20, 2003

### yxgao

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

Thanks so much!!

2. Dec 20, 2003

### Hurkyl

Staff Emeritus
What have you tried so far?

3. Dec 20, 2003

### 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!

4. Dec 20, 2003

### arcnets

yxgao,
here's a hint:
$$e = \sum_{k=0}^\infty \frac{1}{k!}$$
Try to express your sum in terms of this...

5. Dec 20, 2003

### Hurkyl

Staff Emeritus
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.

6. Dec 20, 2003

### 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.

7. Dec 27, 2003

### yxgao

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

8. Dec 27, 2003

### Hurkyl

Staff Emeritus
distribute!

9. Dec 27, 2003

### 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.

10. Dec 27, 2003

### Hurkyl

Staff Emeritus
$$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$$

11. Dec 27, 2003

### yxgao

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

12. Dec 27, 2003

### Hurkyl

Staff Emeritus
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.