GRE 9768 #37

Forum for the GRE subject test in mathematics.
toughluck
Posts: 9
Joined: Thu Jun 25, 2009 10:57 am

GRE 9768 #37

Postby toughluck » Wed Sep 30, 2009 8:33 pm

37. \sum_{k=1}^\infty k^2/k!
(A) e
(B) 2e
(C) (e+1)(e-1)
(D) e^2
(E) infinity

I tried to use the taylor series approximation of e so \sum_{k=0}^\infty x^k/k! and saw that if e is equal to 1 we have \sum_{k=0}^\infty 1/k! but I couldn't figure out what else to do. The answer is b any help would be appreciated.

prong
Posts: 24
Joined: Thu Sep 24, 2009 12:17 am

Re: GRE 9768 #37

Postby prong » Thu Oct 01, 2009 12:23 am

k^2/k! = k/(k-1)! is a useful simplification.

then a hint: sum from 1 to inf over k is the same as sum from 0 to inf over k+1. you can probably get it just from this, i'd advise you not to read the following paragraph until you've tried to work it out using this hint.

use the fact that sum from 1 to inf over k is the same as sum from 0 to inf over k+1. we get sum from 0 to infty of (k+1)/k!. break it up into the two parts. easier part is sum from 0 to infty of 1/k!, you know that's e already. you also have sum from 0 to inf of k/k!, which is 0 + the sum from 1 to inf of k/k!. k/k! = 1/(k-1)! for k > 0 (wouldn't have made sense to do this for 0 included in the sum, since there's nothing to cancel in 0! which is the empty product). now the same trick as before, sum from 1 to infty of 1/(k-1)! = sum from 0 to infty of 1/k!. that's also e.

so it's two parts, each of which is e. therefore 2e.

sjin
Posts: 10
Joined: Sat Sep 19, 2009 5:55 pm

Re: GRE 9768 #37

Postby sjin » Thu Oct 01, 2009 5:28 am

My calculation:


exp(x)= sum(x^k/k!) from k=0 to inf (then take derivative)
exp(x)= sum(k*x^(k-1)/k!) from k=1 to inf (then multiply both sides by x)
x*exp(x)=sum(k*x^k/k!) from k=1 to inf (then take derivative)
x*exp(x)+ exp(x)=sum( k^2*x^(k-1)/k!) from k=1 to inf (then let x=1)
exp+exp==sum(k^2/k!) from k=1 to inf

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: GRE 9768 #37

Postby EugeneKudashev » Tue Apr 06, 2010 1:03 pm

I think one might consider another "GRE-approach" here. that is trying to estimate the answer w/o actually solving the problem. this approach is described thoroughly in MIT course "Street-Fighting Mathematics" (available online here: http://ocw.mit.edu/OcwWeb/Mathematics/1 ... /index.htm ) and is extremely applicable with GRE.

so, let's just write down first few terms of the series:
1 + 4/2 + 9/6 + 16/14 + 25/120 + 36/720 + ...
and try to estimate what's that:
~= 1 + 2 + 1.5 + 0.66 + 0.2 + 0.05 = 5.41.

stop right there and review your possible answers.
A: e~=2.7 < 5.41. out.
B: 2*e~=5.4. very close, let's check other options.
C: (e+1)(e-1)~=6.28 (probably the most time-consuming comp.) - too much. out.
D: e^2~=7.29. out.
E: out as well, we see that denominator increases way faster than numerator (not very rigorous, but works, right?)

so, the answer is B. I believe this might be not very profound solution, however it can save you a priceless minute or two on the real test.

opinions? :)

Andy Chen
Posts: 1
Joined: Fri Aug 06, 2010 12:06 pm

Re: GRE 9768 #37

Postby Andy Chen » Fri Aug 06, 2010 12:13 pm

Use this : k^2/k! = (k-1)/(k-1)! + 1/(k-1)!

Then the summation is immediately know to be 2e.




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