Problem 37 9768

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Problem 37 9768

Postby 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
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby 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
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby 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
Posts: 10
Joined: Mon Oct 13, 2008 10:36 pm

Postby Kastro » Tue Oct 28, 2008 3:21 pm

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




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