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I am required to find the limit of $$ \frac{\sqrt{n}}{2^n}\cdot \binom{n}{\frac{n}{2} + \sqrt{n}} $$

I am given the following hints:

  1. Stirling's approximation $$ n! \approx \sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n $$

  2. For $k \rightarrow \infty$, with $x$ being a constant. $$ \left(1 + \frac{x}{k} \right)^k \rightarrow e^x $$

I've tried doing a direct substitution using Stirling's approximation, but I am stuck with how to further simplify the terms. I am also unsure of how to use (2).

This is the expression I am currently stuck at and have no idea how to proceed: $$ \frac{n^{n+1}}{2^n \cdot \sqrt{2\pi}}\frac{1}{\sqrt{(\frac{n}{2})^2-n}\cdot(\frac{n}{2}-\sqrt{n})^{\frac{n}{2}-\sqrt{n}}\cdot(\frac{n}{2}+\sqrt{n})^{\frac{n}{2}+\sqrt{n}}} $$

Any hints is greatly appreciated.

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  • $\begingroup$ Can you give the expression, as simplified as possible, that you got to? $\endgroup$
    – Tavish
    Apr 29 '21 at 11:08
  • $\begingroup$ Thanks for the reply, I've updated my question to include the expression which I'm stuck at $\endgroup$
    – jibara3179
    Apr 29 '21 at 11:16
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Ignoring the $\sqrt{2\pi}$ factor, you can collect terms as follows: $$2 \cdot\left( \frac{n}{2\sqrt{n^2/4 -n}} \right)^{n+1} \cdot \left( \frac{n/2-\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} $$ Now, the first factor is further simplified to $$\left( \frac{\sqrt n}{\sqrt {n-4}} \right)^{n+1} \\ = \frac{1}{ (1-\frac 4n)^{\frac{n+1}{2}}}\\ = \frac{1}{((1-\frac 4n)^n)^{1/2} \cdot (1-\frac 4n)^{1/2}} $$ $$ \to \frac{1}{(e^{-4})^{1/2}} \cdot \frac{1}{1}\\ = e^2 $$ and the second factor is $$\left(1-\frac{2\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} \\ =\left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n} \\= \left( \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n /2 +1} \right)^2 \cdot \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{-2} \\ \to (e^{-2})^2 \cdot 1 \\ =e^{-4} $$

So, all in all the answer should be $$\frac{1}{\sqrt{2\pi}} \times 2 \times e^2 \times e^{-4}=\color{red}{ \sqrt{\frac{2}{\pi}} e^{-2}} $$

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  • $\begingroup$ Sorry, I'm lost at the first step. I'm not exactly sure how the terms can be collected into $$2 \cdot\left( \frac{n}{2\sqrt{n^2/4 -n}} \right)^{n+1} \cdot \left( \frac{n/2-\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} $$ $\endgroup$
    – jibara3179
    Apr 29 '21 at 12:24
  • $\begingroup$ @jibara3179 Write $\frac{1}{2^n}= 2\cdot \frac{1}{2^{n+1}}$. In the denominator, you have two terms of the form $a^{n/2 -\sqrt n} \cdot b^{n/2 +\sqrt n}$. Separate these as $(ab)^{n/2} \times (b/a)^{\sqrt n} $. Recognize $ab$ as a difference of two squares. Any clearer now? $\endgroup$
    – Tavish
    Apr 29 '21 at 12:34
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    $\begingroup$ Yep, it's clearer now. Thank you. $\endgroup$
    – jibara3179
    Apr 30 '21 at 6:37
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Assuming that you want the limit as $n\rightarrow\infty$.

Hint: One simplifying step you can take: Since $$ \lim_{n\rightarrow\infty}\frac{\frac{n}{2}}{\sqrt{(\frac{n}{2})^2-n}}=1. $$ Therefore, you can replace the radical in your denominator by $\frac{n}{2}$. Other similar simplifications can also be applied to the other factors.

For the second factor in the denominator, $$ (\frac{n}{2}-\sqrt{n})^{\frac{n}{2}-\sqrt{n}}, $$ you can factor out an $n$ to get $$ (\frac{n}{2})^{\frac{n}{2}-\sqrt{n}}(1-\frac{2}{\sqrt{n}})^ {\frac{n}{2}-\sqrt{n}}. $$ This allows you to use the second formula, after working with the exponents.

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