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Erdos remarked somewhere the bound
\[ {{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}. \] This can be established by induction: \[ {{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}} \] and if we have the bound for $n$, we only have to show \[ \frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}} \] which reduces to $4n^2+84+3<4n^2+8n+4$.