Prove that for all positive integer n, the inequality $2n\choose n$ $<4^n$ holds

Question

How do I prove that for all positive integer n, the inequality $2n\choose n$$<4^n$ holds?

Thank you!

See also http://math.stackexchange.com/questions/448861/prove-that-2n-binom2nn-22n and http://math.stackexchange.com/questions/931306/inequality-binom2nn-leq-4n — Martin Sleziak, Oct 04 '16 at 09:13

score 8 · Accepted Answer · answered May 16 '13 at 16:29

8

Hint: The LHS is the number of $n$-element subsets of $[2n]$, while the RHS is the number of all subsets of $[2n]$.

answered May 16 '13 at 16:29

Thanks you! But how do I set the equations up? – Emile May 16 '13 at 16:30
2

@Emile You can use this hint the following way: $4^n=(1+1)^{2n} =\sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n}$. – N. S. May 16 '13 at 16:32
@emile N.S. said it, I was just about to type it, oh by the way, what have you tried? – May 16 '13 at 16:33

1 Answers1