I need some quick help on problem 37 from the practice test:

What is the sum of k^2/k! as k = 0 -> infinity?

The booklet claims the answer is 2*e, and I know for a fact that the sum of 1/k! is e, but I can't logically get from one to the other. I've tried looking at Taylor series, but it's no use. Any help would be greatly appreciated.

## Problem 37 on the Practice Test - I need help.

### The Answer

False alarm, guys. I found the answer on another message board.

First note that, for k >= 1, k^2/k! = k/(k-1)!. By a change of

variable we may write sum(k=1 to oo) k/(k-1)! = sum(k=0 to oo) (k+1)/

k! = sum(k=0 to oo) 1/k! + sum(k=0 to oo) k/k!. The first term is e.

For the second term note that for the case k=0 the summand is zero, so

we can write this as sum(k=1 to oo) k/k! = sum(k=1 to oo) 1/(k-1)! =

sum(k=0 to oo) 1/k!, where the last equality follows from another

change of variable. So this term is also e.

First note that, for k >= 1, k^2/k! = k/(k-1)!. By a change of

variable we may write sum(k=1 to oo) k/(k-1)! = sum(k=0 to oo) (k+1)/

k! = sum(k=0 to oo) 1/k! + sum(k=0 to oo) k/k!. The first term is e.

For the second term note that for the case k=0 the summand is zero, so

we can write this as sum(k=1 to oo) k/k! = sum(k=1 to oo) 1/(k-1)! =

sum(k=0 to oo) 1/k!, where the last equality follows from another

change of variable. So this term is also e.