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 $1$.
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$\begingroup$ Have you tried the Stirling approximation? $\endgroup$ – J.G. Mar 24 '19 at 19:34
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$\begingroup$ Why don't you specify the point in which we must calculate the limit? $\endgroup$ – Eureka Mar 24 '19 at 19:37
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$\begingroup$ Sorry, forgot to specify the point $\endgroup$ – ErlGrey Mar 24 '19 at 19:39
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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.
What I like about this approach is that it doesn't rely on knowing/ proving other results about famous limits.
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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 $ :
$$\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}=0 $$
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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).
