I am attempting to evaluate the limit $\lim_{m \to \infty} \binom{m}{k} \cdot 2^{-m}$ for $0 \leq k \leq m$, and I have been able to confirm via Mathematica that this limit is indeed 0. How would one actually go about proving this?
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Is $k$ fixed, or can it grow with $m$? – Clement C. Mar 07 '21 at 11:23
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It can grow with $m$ too. – Glycerius Mar 07 '21 at 11:24
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1One way then could be to use Stirling's approximation -- the maximum over all $k$ (up to some annoying distinction of cases for $m$ even or odd) if for $k=m/2$, and that one can handled via Stirling's. – Clement C. Mar 07 '21 at 11:27
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For simplicity let $m$ be even. $\binom{m}{k}$ is largest when $k=\frac12m$ [see here]. Hence $$\binom{m}{k}2^{-m}\le\frac{m!}{(m/2)!(m/2)!}2^{-m}.$$ Now using this you can find the limit to be zero.
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Nice! This is quite a clever solution -- that bound for the binomial makes a lot of intuitive sense, but how would one make a rigorous argument for it? – Glycerius Mar 07 '21 at 11:33
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That C(m, k) is largest when k = m / 2. This is obviously true, but what is the rigorous argument for it? – Glycerius Mar 07 '21 at 12:37
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