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 $4n^2+84+3<4n^2+8n+4$.