$\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$ and $\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$

Question

I have two questions:

$$\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$$ and $$\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$$

I realised that I had to do this with the help of integration, and in both cases I had to come to the same situation $\int_{0}^{1}\ln \space xdx$. But then I'm having $(x\times ln\space x-x)|_{0}^{1}$ and am stuck in the $0\times ln \space 0$ thing. What should I do with it?

@DavidMitra, the second answer to that question dectly states $\int_{0}^{1}ln \space xdx=-1$, but I am stuck there. — Indrayudh Roy, Feb 26 '14 at 20:02
See Stirling's approximation and central binomial coefficient — Lucian, Feb 26 '14 at 20:19
For the second, see evaluation of $\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}$ and Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$. — Ben, Feb 27 '14 at 00:23

score 0 · Answer 1 · answered Feb 26 '14 at 21:10

A simple substitution lets you see whats happening. Substitute $x=1/y$ into the limit so that it becomes $$\frac{-ln(y)}{y}.$$ As x approaches zero y approaches infinity, and it's a little more intuitive that the above limit should go to zero. You can prove it with L'Hopital.

$\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$ and $\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$

1 Answers1