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For $k\leq n$, how do I prove that ${2n \choose n}(1-\frac{k}{n})^{k}\leq{2n \choose n+k}$?

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up vote 5 down vote accepted

Hint: If $0\lt a\leqslant b$, then $\dfrac{a}b\leqslant\dfrac{a+1}{b+1}$.

Application: The hint yields $\dfrac{n-k}n\leqslant\dfrac{n-k+i}{n+i}$ for every $i\geqslant0$. Multiplying these inequalities for $1\leqslant i\leqslant k$, one gets $$ \left(\frac{n-k}n\right)^k\leqslant\frac{n-k+1}{n+1}\frac{n-k+2}{n+2}\cdots\frac{n}{n+k}=\frac{n!\,n!}{(n-k)!(n+k)!}. $$ Hence, $$ {2n\choose n}\left(\frac{n-k}n\right)^k\leqslant\frac{(2n)!}{n!\,n!}\,\frac{n!\,n!}{(n-k)!(n+k)!}=\frac{(2n)!}{(n-k)!(n+k)!}={2n\choose n+k}. $$

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Thank you, Didier. –  bscptn Jan 11 '12 at 2:20

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