# Random symmetric walk. [duplicate]

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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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 4 '15 at 14:25

• 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. – Jack D'Aurizio Jun 4 '15 at 12:37