<s>[INST] Proofs from the BOOK: Bertrand's postulate: $\binom{2m+1}{m}\leq 2^{2m}$ I have a very hard proof from "Proofs from the BOOK". It's the section about Bertrand's postulate, page 8:
It's about the part, where the author says:
$$\binom{2m+1}{m}\leq 2^{2m}$$
because $\binom{2m+1}{m}=\binom{2m+1}{m+1}$ are the same in $\sum \limits_{k=0}^{2m+1} \binom{2m+1}{k}=2^{2m+1}$
I see, why they are the same, but I don't see the reason to say $\binom{2m+1}{m}\leq 2^{2m}$. Any help would be fine :)
 [/INST] Since $2m+1$ is odd, there will be an even number of coefficients (since you have everything from $\binom{2m+1}{0}$ to $\binom{2m+1}{2m+1}$). Because the coefficients are symmetric, you can write:
$\sum \limits_{k=0}^{m} \binom{2m+1}{k}=\sum \limits_{k=m+1}^{2m+1} \binom{2m+1}{k}$
But $\sum \limits_{k=0}^{2m+1} \binom{2m+1}{k}=\sum \limits_{k=0}^{m} \binom{2m+1}{k} + \sum \limits_{k=m+1}^{2m+1} \binom{2m+1}{k}=2^{2m+1}$
$2 \times \sum \limits_{k=0}^{m} \binom{2m+1}{k} = 2^{2m+1}$
$\sum \limits_{k=0}^{m} \binom{2m+1}{k} = 2^{2m}$
Since $\binom{2m+1}{m}$ is part of the sum $\sum \limits_{k=0}^{m} \binom{2m+1}{k}$, it follows that $\binom{2m+1}{m}\leq 2^{2m}$, equality when $m=0$
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