Problem 37 9768

moo5003 · Post by **moo5003** » Mon Oct 27, 2008 5:19 pm

37. What is the summation from 1 to Infinity of k^2/k!

A) e
B) 2e
C) (e+1)(e-1)
D) e^2
E) Infinity

Answer - B

So, I tried expanding e with its taylor series namely
1/k! and then somehow incorporate k^2 on top with no avail. I'm a little unsure how the answer is 2e if anyone could explain this to me I would appreciate it.

I was under the assumption 2e = Summation of 2/k!

Nameless · Post by **Nameless** » Mon Oct 27, 2008 8:06 pm

Hi
the problem was solved, the idea is :
k^2/k!=k/(k-1)! =(k-1+1)/(k-1)!=1/(k-2)! +1/(k-1)!

Q.E.D

moo5003 · Post by **moo5003** » Tue Oct 28, 2008 1:40 pm

Isnt it undefined to have 1/(k-2)! since the sum is from 1 to infinity and for k=1 you need to evaluate (1-2)! = (-1)!

Kastro · Post by **Kastro** » Tue Oct 28, 2008 3:21 pm

Yep. Throw out those terms and start the summation later, basically.