Forum for the GRE subject test in mathematics.

moo5003
 Posts: 36
 Joined: Mon Oct 06, 2008 7:33 pm
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)(e1)
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
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
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by Nameless » Mon Oct 27, 2008 8:06 pm
Hi
the problem was solved, the idea is :
k^2/k!=k/(k1)! =(k1+1)/(k1)!=1/(k2)! +1/(k1)!
Q.E.D

moo5003
 Posts: 36
 Joined: Mon Oct 06, 2008 7:33 pm
Post
by moo5003 » Tue Oct 28, 2008 1:40 pm
Isnt it undefined to have 1/(k2)! since the sum is from 1 to infinity and for k=1 you need to evaluate (12)! = (1)!

Kastro
 Posts: 10
 Joined: Mon Oct 13, 2008 10:36 pm
Post
by Kastro » Tue Oct 28, 2008 3:21 pm
Yep. Throw out those terms and start the summation later, basically.