Question

calculus question! prove that lim n-> infinity (2^n n!)/n^n = 0.

Answer 1

\[\lim_{n \rightarrow \infty} 2^n n!\div n^n =0\]

Answer 2

prove\[\lim_{n\to\infty}{2^nn!\over n^n}=0\]ha, and you have another good problem going as well, why not focus on that one first while I feebly attempt this one

Answer 3

log tricks?

Answer 4

show

\[\sum_{n=1}^{\infty}(2^n n!)/n^n\] converges

Answer 5

root test then?

Answer 6

ratio test works nice

Answer 7

can you show me the steps? :0

Answer 8

it is also easy to get an upper bound for (2^n n!)/n^n that also goes to zero

Answer 9

\[a_n={2^nn!\over n^n}\]\[a_{n+1}={2^{n+1}(n+1)!\over (n+1)^{n+1}}\]\[\large \lim_{n\to\infty}|{a_{n+1}\over a_n}|=\lim_{n\to\infty}\frac{2n^n}{(n+1)^n}=2e^{\lim_{n\to\infty}n\ln{n\over n+1}}\]just looking at the exponent and using log properties we have\[\lim_{n\to\infty}n\ln{n\over n+1}=...?\]@Zarkon what would you do now? n=1/t ?

Answer 10

\[\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n\]
\[=\left(\frac{n+1-1}{n+1}\right)^n=\left(1-\frac{1}{n+1}\right)^n\]
\[=\left(1-\frac{1}{n+1}\right)^{n+1}\left(1-\frac{1}{n+1}\right)^{-1}\]
\[\to e^{-1}\cdot 1=e^{-1} \text{ as }n\to\infty\]

Answer 11

oh that's just so clever it makes me sick ;)
Thanks you master Zarkon!

Answer 12

I guess we should point out to the OP that if you have a series that converges then the sequence converges to zero. Thus we have shown that \(\displaystyle \lim_{n\to\infty}{2^nn!\over n^n}=0\)

Answer 13

\[(\frac{n}{n+1})^n=(\frac{n+1}{n})^{-n}=(1+\frac{1}{n})^{-n}=e^{-1}\] just thought i would put my two cents in