We have the sequence $$a_n=\frac{(2n)!^2}{16^n\cdot n!^4}$$ I am trying to prove this converges to 0, which is relatively easy using Stirling's Approximation (which I don't want to use). I have tried a myriad of things so far, what I think is the closest thing I have to a solution is rewriting it as a power series: $$a_n=\prod_{k=1}^n\frac{(k+n)^2}{16k^2}$$ From here we can see that for all $k$ greater than $k_0=\frac{1}{3}n$ the terms of the power series are smaller than 1, but I don"t know if this is enough to prove anything. Any help is much appreciated.
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1The equivalent of $\sqrt{a_n}$ as $n\to\infty$ is a classic application of Wallis integrals. (it's a way to prove Stirling) – Jean-Claude Arbaut Nov 27 '22 at 21:04
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1By the way, you haven't written a power series—it's a finite product. – Greg Martin Nov 27 '22 at 21:33
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@GregMartin but isn't it a power series for $\lim_{n\to\infty}$? – John Doe Nov 27 '22 at 21:35
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In the title you're asking about a series but in the main text it's about a sequence. Which is it? – lcv Nov 28 '22 at 08:55
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$$\small 0<a_n=\left(\frac12\cdot\frac34\cdots\frac{2n-1}{2n}\right)^2<\frac12\cdot\frac34\cdots\frac{2n-1}{2n}\cdot\frac23\cdot\frac45\cdots\frac{2n}{2n+1}=\frac1{2n+1}\underset{n\to\infty}{\longrightarrow}0.$$
metamorphy
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