# Find the limit $\lim_{n\rightarrow\infty}\frac{(100n)!}{(99n)!(100n)^n}$

Find the limit $$\lim_{n\rightarrow\infty}\frac{(100n)!}{(99n)!(100n)^n}$$

How to prove that this limit is $$0$$? I've tried to find it's upper estimate that tends to zero, but did not find something better than with limit $$1$$.

• Have you tried the Stirling approximation?
– J.G.
Mar 24, 2019 at 19:34
• Why don't you specify the point in which we must calculate the limit? Mar 24, 2019 at 19:37
• Sorry, forgot to specify the point Mar 24, 2019 at 19:39

Note that:

$$\frac{100n!}{99n!(100n)^n} = \frac{(99n+1)(99n+2)\ldots (99n+n-1)(100n)}{(100n)^n} = \prod_{k=1}^{n} \frac{99n+k}{100n} \leq \Big(\frac{99n+n/2}{100n}\Big)^{n/2}$$ where the inequality at the end comes from using 1 as an upper estimate for the last $$n/2$$ terms and $$\Big(\frac{99n+n/2}{100n}\Big)$$ as an upper estimate for the first $$n/2$$ terms. By canceling the $$n$$'s inside this is just equal to $$(99.5/100)^{n/2}$$ which clearly goes to 0.

You could use Stirling approximation for $$x \to \infty$$:

$$x!\approx \sqrt{2\pi}\frac{x^{x+\frac{1}{2}}}{e^x}$$

So the limit becomes:

$$\lim_{x \to \infty} \frac{\sqrt{2\pi}\frac{(100x)^{100x+\frac{1}{2}}}{e^x}}{\sqrt{2\pi}\frac{(99x)^{100x+\frac{1}{2}}}{e^x}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{100x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{99x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}}=$$ $$=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(100x)^{99x}}{(99x)^{100x}}=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}$$

For infinites hierarchy $$x^x>>a^x$$ :

$$\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}=0$$

Let $$a_n=\frac{(100n)!}{(99n)!(100n)^n}$$ and evaluate the limit of the ratio \begin{align}\frac{a_{n+1}}{a_n}&=\frac{(100(n+1))!}{(99(n+1))!(100(n+1))^{n+1}}\cdot \frac{(99n)!(100n)^n}{(100n)!}\\ &=\frac{(100n+99)\cdots (100n+1)}{(99n+99)\cdots (99n+1)(1+\frac{1}{n})^n}\to \frac{(1+\frac{1}{99})^{99}}{e}<1.\end{align} Then apply the ratio test for sequences (see Proof attempt to the ratio test for sequences).