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I am stuck on the following problem to find the limit of the sequence $a_n=\frac{(n!)^2 \cdot (2n)!}{(4n)!}$. I show $\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| < 1$, thus $a_n \rightarrow 0$. However, I would like to figure out a more algebraic way of doing this, although I am stuck on how to simplify this expression. Could someone offer a hint on how to get started, not a full solution please.

Thank you.

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  • $\begingroup$ Hint: Stirling's approximation should help you get rid of those factorials and get more convenient expressions. $\endgroup$
    – TMM
    May 10, 2017 at 15:02
  • $\begingroup$ (Note: it is an approximation, but for large n it is a tight approximation; the relative error is negligible. So it will give you exact results regarding the asymptotic scaling of a_n, even if it is an "approximation".) $\endgroup$
    – TMM
    May 10, 2017 at 15:07

2 Answers 2

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At the request of the OP, we present herein a purely algebraic way forward.

HINT:

Let $a_n=\frac{(n!)^2\,(2n)!}{(4n)!}$. Then, we can write

$$\begin{align}\log(a_n)&=2\log(n!)+\log((2n)!)-\log((4n)!)\\\\&=2\sum_{k=1}^n\log(k) +\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{2n+1}^{4n}\log(k)\end{align}$$

SPOLIER ALERT: Scroll Over the Highlighted Area to Reveal the Full Solution

\begin{align}\log(a_n)&=2\log(n!)+\log((2n)!)-\log((4n)!)\\\\&=2\sum_{k=1}^n\log(k) +\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{2n+1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{k=1}^{2n}\log(k+2n)\\\\&\le 2n\log(n)-2n\log(2n+1)\\\\&=-2n\log\left(2+\frac1n\right)\\\\&<-2n\end{align}Inasmuch as $\lim_{n\to \infty}e^{-2n}=0$, we find that $$\lim_{n\to \infty}a_n=0$$

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  • $\begingroup$ Thank you for your solution, this clears my question up and also gives me "tricks" for solving such problems in the future +1, and accepted answer. $\endgroup$
    – Joe
    May 11, 2017 at 1:39
  • $\begingroup$ You're welcome Joseph! My pleasure. And thank you for accepting the answer. -Mark $\endgroup$
    – Mark Viola
    May 11, 2017 at 1:47
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Start with this and simplify $$ \frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2 \cdot (2(n+1))!}{(4(n+1))!} \cdot \frac{(4n)!}{(n!)^2 \cdot (2n)!}\\ = \frac{((n+1)n!)^2 \cdot (2n+2)(2n+1)(2n)!}{(4n+4)(4n+3)(4n+2)(4n+1)(4n)!} \cdot \frac{(4n)!}{(n!)^2 \cdot (2n)!} $$

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  • $\begingroup$ @Ihf yes, my mistake I fixed it. $\frac{a_{n+1}}{a_n}=\frac{n^2+2n+1}{64n^2+64n+12} \rightarrow \frac{1}{64}$ as $n \rightarrow \infty$. I used this fact to show $L < 1 \rightarrow a_n \rightarrow 0$ $\endgroup$
    – Joe
    May 10, 2017 at 16:06
  • $\begingroup$ @Joseph, exactly! $\endgroup$
    – lhf
    May 10, 2017 at 16:38

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