## Sequences, series

Forum for the GRE subject test in mathematics.
CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

### Sequences, series

What is
lim_{n->\infty}{(n!)^(1/n)}?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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.

Posts: 24
Joined: Sun Oct 05, 2008 1:41 am
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
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
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
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

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
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
in I is m a constant or a typo?
typo
corrected
sorry!

Posts: 24
Joined: Sun Oct 05, 2008 1:41 am
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
mistake in my last problem also - corrected now - sorry

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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)

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Hey friends, can we embed LATEX into this site ?

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
According to $n!\ ge (\frac{n+1}3)^n$ we get $lim_{n\to \infty}(n!)^1/n\to \infty$