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So there's this assignment I'm doing. Let p=1/2. I already proved that for a random walk P(X_n=k) = (n over (n+k)/2) * 2^(-n)

Now I need to prove that lim n->inf (n^(1/2))P(X_2n=2k)) = 1/pi.

Given is stirling's limit. Though I have no idea how to do this.

I'm typing this from my phone and I have no idea how to write in latex, so sorry.


marked as duplicate by Jack D'Aurizio, user147263, Did limits Jun 4 '15 at 14:25

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  • $\begingroup$ You have to prove $$\lim_{n\to +\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{\pi}}$$ that is trivial through Stirling's formula. $\endgroup$ – Jack D'Aurizio Jun 4 '15 at 12:37
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Jun 4 '15 at 12:53
  • $\begingroup$ The answer by Ragib Zaman there ^ $\endgroup$ – user147263 Jun 4 '15 at 13:32