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I'm asked to prove that $$\lim_{n \to \infty}\Bigl (\frac{n^n}{(2n)!} \Bigr )=0$$ I thought about using Stirling's approximation since $n \rightarrow\infty$ and then L'hopital in order to prove it directly (Stirling's approximation in order to avoid deriving factorial, which I'm not supposed to know how, given the fact it envolves gamma function). But then I get this: $$\lim_{n \to \infty}\Bigl (\frac{n^n}{(2n)!} \Bigr )=\lim_{n \to \infty}\Bigl (\frac{n^n}{2n\cdot ln(2n)-2n} \Bigr ) = \lim_{n \to \infty}\Bigl (\frac{n^{n-1}}{2ln(2n)-2} \Bigr )="\frac{\infty}{\infty}"$$ Therefore by L'hopital: $$\lim_{n \to \infty} \frac{n^{n-1}}{2ln(2n)-2}=\lim_{n \to \infty} \frac{n^{n-1}\cdot ln(n-1)}{2(\frac{2}{2n})}=\lim_{n \to \infty} \frac{n^{n}\cdot ln(n-1)}{2} \rightarrow\infty$$ My direct approach of proof failed, and I dont come up with a way other than doing the following:

I thoght about simplifying this, and saying I have $n$ factors on my numerator and $2n$ factors on my denominator, meaning: $$\frac{n\cdot n\cdot n....\cdot n}{1\cdot 2\cdot3\cdot...\cdot n\cdot(n+1)\cdot...\cdot(2n-1)\cdot(2n)}$$ Now if I diminish 1 $n$ from both the numerator and denomenator I'm left with $(n-1)$ times $n$ and both $(n-1)$ factors that are smaller than $n$ and $n$ factors bigger than $n$. Therefore my claim is that the overall product in the denomenator shour be bigger, meaning:$$\Bigl(\prod_{i=1}^{n-1}i \Bigr)\times\Bigl(\prod_{i=n+1}^{2n}i \Bigr)\geq \Bigl(\prod_{n=1}^{n-1}n \Bigr)$$

Is that claim too ligical but not enough mathematical, or it should be enough to justify my proof? thanks a lot in advance.

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  • $\begingroup$ In your first line, obviously $(2n)! \not \sim 2n \log(2n) - 2n$. $\endgroup$ – Qi Zhu Apr 9 at 11:29
  • $\begingroup$ on high values of n, the basic assumption of Stirling's Approximation is that equation is true. $\endgroup$ – Avi P Apr 11 at 7:29
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    $\begingroup$ No! Of course not! Look up Stirling again. You should develop some intuition for the growth rate of the functions. Obviously, the left side is much much much larger than the right for large $n$. $\endgroup$ – Qi Zhu Apr 11 at 7:49
  • $\begingroup$ In a sense, I'm trying to develop that intuition. thanks for the insight :) $\endgroup$ – Avi P Apr 11 at 10:24
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$n+i ≥ n$ $\forall i \in \mathbb{N}$

Then $(n+1)...(2n)≥n^{n}$

Then $1≥ \frac{n^{n}}{(n+1)...(2n)}$

Then $\frac{1}{n!}≥\frac{n^{n}}{(2n)!}$

Then $0≤\lim_{n \to \infty} (\frac{n^{n}}{(2n)!})≤ \lim_{n \to \infty} (\frac{1}{n!}) = 0$

And we are done!

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You may just write

$$\frac{n^n}{(2n)!}= \frac{1}{n!}\cdot \frac{n^n}{(n+1) \cdots (2n-1)(2n)} \leq \frac{1}{n!} \stackrel{n \to \infty}{\longrightarrow}0$$

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  • $\begingroup$ I guess that works too,haha. thank you! $\endgroup$ – Avi P Apr 9 at 11:22
  • $\begingroup$ @AviP : You are welcome :-) $\endgroup$ – trancelocation Apr 9 at 11:25
  • $\begingroup$ By the way... any guess why Stirling approximation + L'hopital failed in this case? $\endgroup$ – Avi P Apr 9 at 11:27
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Alternatively, by ratio test, the series $$\sum_n \frac{n^n}{(2n)!} < \infty$$

Since $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(2n+2)!} \times\frac{(2n)!}{n^n} $$

$$\hspace{4.2cm} =\left(1+\frac{1}{n}\right)^n \frac{(n+1)}{(2n+2)(2n+1)} \rightarrow 0<1 $$so its $n$-th term tend to zero!

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