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I am trying to show the folllowing property:

$\binom{2m}{m} \leq 2^{2m} \frac{2m-1}{6m} \ \ \ \ \ $ (for $m\geq 4$)

I first tried using the known result: $\binom{n}{k} \leq (\frac{ne}{k})^k$ however it doesn't seem to work. I'm stuck at this point.

Any help would be appreciated

DIexp
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  • Have you tried induction? – GReyes Apr 12 '23 at 14:37
  • See also https://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates and https://mathoverflow.net/questions/133732/upper-limit-on-the-central-binomial-coefficient – lhf Apr 12 '23 at 14:42
  • @lhf - thanks! :) – DIexp Apr 12 '23 at 14:56
  • You should not expect $\binom nk \le \frac{n^k}{k!} \le (\frac{ne}{k})^k$ to work well when $k$ is linear in $n$. Even when $k$ is around $\sqrt n$ it might not always be good enough. – Misha Lavrov Apr 12 '23 at 15:18

1 Answers1

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This can be proved by induction.

If $m=4$, then $\binom{2m}{m}=70$ and $2^{2m}\frac{2m-1}{6m}=\frac{224}{3}$. And $70=\frac{210}3<\frac{224}{3}$.

Now, take $m\in\mathbb{N}$ and assume that $$ \binom{2m}m\le2^{2m}\frac{2m-1}{6m}.\label{ineq}\tag1 $$Then, since $$ \binom{2(m+1)}{m+1}=\binom{2m}m\times\frac{4m+2}{m+1} $$ and $$2^{2(m+1)}\frac{2(m+1)-1}{6(m+1)}=2^{2m}\frac{2m-1}{6m}\times\frac{4m(2m+1)}{(m+1)(2m-1)} $$ then, in order to deduce from \eqref{ineq} that $$ \binom{2(m+1)}{m+1}\le2^{2(m+1)}\frac{2(m+1)-1}{6(m+1)}, $$ it is enough to prove that $$ \frac{4m+2}{m+1}\le\frac{4m(2m+1)}{(m+1)(2m-1)}. $$ But we have \begin{align} \frac{4m+2}{m+1}\le\frac{4m(2m+1)}{(m+1)(2m-1)}&\iff2(2m+1)\le\frac{4m(2m+1)}{2m-1}\\ &\iff2\le\frac{4m}{2m-1}\\ &\iff4m-2\le4m, \end{align} and this is a true statement.

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