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Evaluate $\lim_{n \to +\infty} \frac{(2n)!\sqrt n}{2^{2n}\cdot (n!)^{2}}$.

Please help with steps, Dont know how to break it down to cancel out terms.

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    $\begingroup$ This limit diverges to positive infinity $\endgroup$ – Peter Foreman Feb 4 at 18:58
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    $\begingroup$ Do you know of Stirling's approximation? $\endgroup$ – user170231 Feb 4 at 19:00
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    $\begingroup$ Welcome to Maths SX! Could you write the formula with MathJax, so there's no ambiguity? $\endgroup$ – Bernard Feb 4 at 19:01
  • $\begingroup$ the equation is correct now $\endgroup$ – Luis Feb 4 at 20:38
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Hint: using the Stirling approximation:$$ n!=\left(\frac n e\right)^n\sqrt{2\pi n}$$ one easily finds that the limit is$$ \frac1{\sqrt\pi}.$$

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