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How would you prove $ \binom{2n}{n} $ ~ $ \frac{4^n}{\sqrt{πn}} $

Using Stirling's formula: $ n! $ ~ $ (\frac{n}{e})^n\sqrt{2πn} $

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    $\binom{2n}{n}$ is `\binom{2n}{n}`. :) – 2017-02-20
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    Do you have any ideans regarding this problem? – 2017-02-20
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    Do you know how to express $\binom{2n}{n}$ as a product of factorials? – 2017-02-20

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$$C_{n}^{2n}=\frac{2n!}{n!n!}\sim\frac{(\frac{2n}{e})^{2n}\sqrt{4\pi{n}}}{(\frac{n}{e})^{2n}2\pi{n}}=\frac{4^n}{\sqrt{\pi{n}}}$$

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    Sorry can you explain how you got this? Like what theorems you used? – 2017-02-21
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    I used the Stirling's formula) – 2017-02-21
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    For the last step, is it just plain simplifying? I'm having trouble understanding how you got from the 3rd to lest simplification – 2017-02-21
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    Yes)) simply cut the fraction) – 2017-02-21