How to show that $\sum _{m=1}^{n/4}{\textstyle (}\genfrac{}{}{0ex}{}{n}{m}{\textstyle )}{2}^{-n}\le \mathrm{exp}(-n/8)$

pouzdrotf
2022-07-08
Answered

How to show that $\sum _{m=1}^{n/4}{\textstyle (}\genfrac{}{}{0ex}{}{n}{m}{\textstyle )}{2}^{-n}\le \mathrm{exp}(-n/8)$

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Ordettyreomqu

Answered 2022-07-09
Author has **22** answers

For convenience, lets replace $n\to 4n$, so that it is equivalent to show $\sum _{m=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m}{\textstyle )}\u2a7d{2}^{4n}{e}^{-n/2}$. We need a good bound for the partial sum in LHS, so consider

$\begin{array}{}& \$">\text{(for}mn\text{)}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m-1}{\textstyle )}=\frac{(4n)!}{m!(4n-m)!}\cdot \frac{m}{4n-m+1}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m}{\textstyle )}\cdot \frac{1}{3}\end{array}$

As $1+\frac{1}{3}+\frac{1}{{3}^{2}}+\cdots =\frac{3}{2}$, we get the bound $\sum _{m=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m}{\textstyle )}<\frac{3}{2}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{n}{\textstyle )}$ for the partial sum. Now, using an easily found and rather well known upper bound for binomial coefficients, viz. $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}<\frac{{n}^{k}}{k!$, it is enough to show

$\frac{3}{2}\frac{(4n{)}^{n}}{n!}<{2}^{4n}\cdot {e}^{-n/2}$

which is straightforward by induction.

$\begin{array}{}& \$">\text{(for}mn\text{)}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m-1}{\textstyle )}=\frac{(4n)!}{m!(4n-m)!}\cdot \frac{m}{4n-m+1}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m}{\textstyle )}\cdot \frac{1}{3}\end{array}$

As $1+\frac{1}{3}+\frac{1}{{3}^{2}}+\cdots =\frac{3}{2}$, we get the bound $\sum _{m=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{m}{\textstyle )}<\frac{3}{2}{\textstyle (}\genfrac{}{}{0ex}{}{4n}{n}{\textstyle )}$ for the partial sum. Now, using an easily found and rather well known upper bound for binomial coefficients, viz. $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}<\frac{{n}^{k}}{k!$, it is enough to show

$\frac{3}{2}\frac{(4n{)}^{n}}{n!}<{2}^{4n}\cdot {e}^{-n/2}$

which is straightforward by induction.

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