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How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$
I don't have idea and I will be happy for help.

AskMath
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    https://math.meta.stackexchange.com/a/27933/44121 – Jack D'Aurizio Apr 14 '18 at 18:02
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    Are you aware that $\sqrt{n+1}-\sqrt{n}\sim\frac{1}{2\sqrt{n}}$ and that $$\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$$ as $n\to +\infty$? Simple manipulations give that the limit equals $\frac{1}{2}$. – Jack D'Aurizio Apr 14 '18 at 18:03
  • @JackD'Aurizio Ok, but what you advice that I will do? I really don't know how to going about it. – AskMath Apr 14 '18 at 18:05
  • @JackD'Aurizio It's not so clear what you did. – AskMath Apr 14 '18 at 18:08
  • Well, think about it and improve your question by adding an attempt, even partial. – Jack D'Aurizio Apr 14 '18 at 18:19
  • my problem is to simplify $\frac{(\frac{n+1}{2})!}{(\frac{n}{2})!}$.. – AskMath Apr 14 '18 at 18:21
  • @askmath Do you know how the factorial is defined for non-integer arguments? Are you familiar with Stirling's formula? Apply it here and turn the crank. – Mark Viola Apr 14 '18 at 18:29
  • @AskMath Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 19 '18 at 19:18

2 Answers2

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Let apply Stirling’s approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

that is

$$\left[\left(\frac{n+1}{2}\right)!\right]^2\sim 2 \pi \frac{n+1}2\left(\frac{n+1}{2e}\right)^{n+1}=\pi\frac{(n+1)^{n+2}}{2^{n+1}e^{n+1}}$$

$$\left[\left(\frac{n}{2}\right)!\right]^2\sim 2 \pi \frac n 2\left(\frac{n}{2e}\right)^n=\pi\frac{n^{n+1}}{2^ne^n}$$

and

$$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}\sim \frac1{2\sqrt n}$$

then

$$\frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}\sim 7^{\frac1{2\sqrt n}}\frac{(n+1)^{n+1}}{2^{n+1}e^{n+1}}\frac{2^ne^n}{n^{n+1}}=\frac{7^{\frac1{2\sqrt n}}}{2e}\left(1+\frac1n\right)^{n+1}$$

user
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For any $n\in\mathbb{N}$

$$ \frac{1}{4^n}\binom{2n}{n} = \frac{\Gamma(2n+1)}{4^n \Gamma(n+1)^2} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \approx \frac{2}{\pi}\int_{0}^{\pi/2} e^{-n\theta^2}\,d\theta \approx \frac{2}{\pi}\int_{0}^{+\infty} e^{-n\theta^2}\,d\theta $$ and the RHS equals $\frac{1}{\sqrt{\pi n}}$. Actually both the $\approx$ above hold as $\sim$ as $n\to +\infty$, and the chain of inequalities still holds if $n\in\frac{1}{2}+\mathbb{N}$. In particular, by letting $n=m/2$ we have $$ \frac{m!}{\left(\frac{m}{2}!\right)^2} \sim \frac{2^{m+1/2}}{\sqrt{\pi m}} $$ hence $$ \frac{\left(\frac{n+1}{2}!\right)^2}{\left(\frac{n}{2}!\right)^2}\sim\frac{2^{n+1/2}}{n!\sqrt{\pi n}}\cdot\frac{(n+1)!\sqrt{\pi(n+1)}}{2^{n+3/2}}\sim\frac{n+1}{2}$$ and the given limit equals $$ \lim_{n\to +\infty}\frac{7^{\sqrt{n+1}-\sqrt{n}}}{2}=\frac{1}{2}.$$ Stirling's approximation is not needed.

Jack D'Aurizio
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