Question

Evaluate lim n->infinity ((n!)^(1/n))/n .
Hint: Compare the logarithm of the above expression with the definition of the integral as 0->1 ln(x) dx

Answer 1

\[Evaluate \lim_{n \rightarrow \infty}(n!)^{1/n}\div n. \]

Answer 2

Hint: Compare the logarithm of the above expression with the definition of the integral \[\int\limits_{0}^{1} \ln(x) dx\]

Answer 3

I'm not sure about the integral, but maybe you can let y = the limit, take the natural log of both sides, then apply L'hopital's?

Answer 4

can you show me? its calculus 2..

Answer 5

\[y = \lim_{n \rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}\]\[\ln y = \lim_{n \rightarrow \infty} \frac {1}{n} \ln \frac {(n!)}{n} = \lim_{n \rightarrow \infty} \ln \frac {(n!)}{n^2}\]

Answer 6

Not sure what to do after that :(
I gotta go outside for a bit, I'll try it again when I get back.

Answer 7

\[\frac{n!^{1/n}}{n}=\frac{(1\cdot2\cdots(n-1)n)^{1/n}}{n}=\frac{\displaystyle\prod_{i=1}^{n}i^{1/n}}{n}\]

\[L=\lim_{n\to\infty}\frac{\displaystyle\prod_{i=1}^{n}i^{1/n}}{n}\]


\[\ln(L)=\lim_{n\to\infty}\ln\left(\frac{\displaystyle\prod_{i=1}^{n}i^{1/n}}{n}\right)\]

\[\ln(L)=\lim_{n\to\infty}\ln\left(\displaystyle\prod_{i=1}^{n}i^{1/n}\right)-\ln(n)\]


\[\ln(L)=\lim_{n\to\infty}\left[\sum_{i=1}^{n}\ln\left(\displaystyle i^{1/n}\right)\right]-\ln(n)\]


\[\ln(L)=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{n}\ln\left(\displaystyle i\right)\right]-\ln(n)\]

\[\ln(L)=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{n}\ln\left(\displaystyle i\right)\right]-\frac{1}{n}\left[\sum_{i=1}^{n}\ln(n)\right]\]

\[\ln(L)=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{n}\ln\left(\displaystyle i\right)-\ln(n)\right]\]

\[\ln(L)=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{n}\ln\left(\displaystyle i/n\right)\right]\]

\[\ln(L)=\int\limits_{0}^{1}\ln(x)dx=-1\]

thus

\[L=e^{-1}=\frac{1}{e}\]

Answer 8

thank u so much can you answer my other question too?:)