Forum for the GRE subject test in mathematics.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Tue Oct 14, 2008 5:07 pm
What is
lim_{n>\infty}{(n!)^(1/n)}?

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
Post
by Nameless » Tue Oct 14, 2008 6:14 pm
Use the Sterling formula : lim (n!/[sqrt(2pi * n)e^(n)n^(n)])=1 when n goes to infinity
so the answer will be infinity
correct me if I am wrong
Last edited by
Nameless on Tue Oct 14, 2008 6:34 pm, edited 1 time in total.

EmLasker
 Posts: 24
 Joined: Sun Oct 05, 2008 1:41 am
Post
by EmLasker » Tue Oct 14, 2008 6:33 pm
I think you are correct nameless...
If we recall that
e = lim (n/(n!^(1/n))), then lim (n!^(1/n)) = lim [(n!^(1/n))/n]*n = (1/e)*(lim n) = infinity.

Kastro
 Posts: 10
 Joined: Mon Oct 13, 2008 10:36 pm
Post
by Kastro » Tue Oct 14, 2008 7:20 pm
We know that n! increases more rapidly than k^n, where k is constant. That is, for all values of k there exists an N such that n > N implies n! > k^n
But then we must have that n!^(1/n) > k
Since k is unbound, n!^(1/n) is unbound above, implying that the limit must go to infinity.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Tue Oct 14, 2008 9:55 pm
Consider 3 series
\sum_{n=1}^{infinity}{a_n}, with each a_n given below. Which of them converge(s)?
I. a_n =
{log(n^{2})} / (n^{2})
II. a_n =
(log 4) / (2n)
III. a_n =
n / (2^n)
A. None
B. I only
C. II only
D. III only
E. More than one of the series converge.
Last edited by
CoCoA on Tue Oct 14, 2008 10:47 pm, edited 1 time in total.

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
Post
by Nameless » Tue Oct 14, 2008 10:41 pm
The problem is straightforward :
I ) diverges since lim (a_n) !=0
II) = log(2)* harmonic series , hence deverges
III) use the ratio test lim(a_n+1/ a_n)=1/2 so the series converges
Therefore the answer is D
Last edited by
Nameless on Tue Oct 14, 2008 10:42 pm, edited 1 time in total.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Tue Oct 14, 2008 10:41 pm
Evaluate sum from n=0 to infinity of
{(1)^n} / {(2n+1)(3^n)}:
A. 1/(2e^3)
B. {e^(1/3) }/2
C. (3^{1/2)*pi)/6
D. 3/(pi*{1/2})
E. Diverges
Last edited by
CoCoA on Tue Oct 14, 2008 10:49 pm, edited 1 time in total.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Tue Oct 14, 2008 10:45 pm
in I is m a constant or a typo?
typo
corrected
sorry!

EmLasker
 Posts: 24
 Joined: Sun Oct 05, 2008 1:41 am
Post
by EmLasker » Tue Oct 14, 2008 10:49 pm
How about one from the 05 practice test...
Find the set of real numbers for which the series converges:
Sum[1 to inf] [n!*x^(2n)]/[n^n(1+x^(2n))]

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Tue Oct 14, 2008 10:51 pm
mistake in my last problem also  corrected now  sorry

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
Post
by Nameless » Tue Oct 14, 2008 10:55 pm
since X^2n>=0 and X^2n/(1+X^2n)<1 so
the series <=sum(n=1..infinity)n!/n^n
Use the Sterling formula : lim (n!/[sqrt(2pi * n)e^(n)n^(n)])=1 when n goes to infinity and the ROOT test, then the radius of convergence is infinity
the answer is e) R

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
Post
by Nameless » Wed Oct 15, 2008 6:01 pm
Evaluate sum from n=0 to infinity of
{(1)^n} / {(2n+1)(3^n)}:
A. 1/(2e^3)
B. {e^(1/3) }/2
C. (3^{1/2)*pi)/6
D. 3/(pi*{1/2})
E. Diverges
The series is convergence so eliminate E)
We have 1/(1+x^2)= sum(n=0...infinity)[(1)^n]x^(2n)
take the integral from 0 to 1/sqrt(3) both sides then the answer is C)

Hey friends, can we embed LATEX into this site ?

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
Post
by CoCoA » Wed Oct 15, 2008 7:56 pm
Correct  Outstanding work. You are ready for your exam!

freddie
 Posts: 2
 Joined: Mon Oct 27, 2008 5:24 am
Post
by freddie » Mon Oct 27, 2008 6:01 am
According to $n!\ ge (\frac{n+1}3)^n$ we get $lim_{n\to \infty}(n!)^1/n\to \infty$