What is

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

What is

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

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

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.

\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.

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

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.

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 ?

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