Calculate the sum $1+\frac{\binom{2}{1}}{8}+\frac{\binom{4}{2}}{8^2}+\frac{\binom{6}{3}}{8^3}+...+\frac{\binom{2n}{n}}{8^n}+...$
Problem
Source: III International Festival of Young Mathematicians Sozopol 2012, Theme for 10-12 grade
Tags: algebra, binomial sum, Sum, Calculate
pco
26.11.2019 17:46
Pinko wrote: Calculate the sum $1+\frac{\binom{2}{1}}{8}+\frac{\binom{4}{2}}{8^2}+\frac{\binom{6}{3}}{8^3}+...+\frac{\binom{2n}{n}}{8^n}+...$ Note that with $f(x)=(1-x)^{-\frac 12}$, easy induction gives n-derivative $f^{(n)}(x)=4^{-n}\frac{(2n)!}{n!}(1-x)^{-\frac{2n+1}2}$ And so taylor development of $f(x)$ at zero is $(1-x)^{-\frac 12}=\sum_{n=0}^{\infty}\left(\frac x4\right)^n\binom{2n}n$ And so, setting $x=\frac 12$ : $\boxed{\sum_{n=0}^{\infty}\frac{\binom{2n}n}{8^n}=(1-\frac 12)^{-\frac 12}=\sqrt 2}$
Hexagrammum16
26.11.2019 17:51
$\sqrt{2}$
$\binom{2n}{n} = \frac {2^n \cdot 1 \cdot 3 \cdot 5 \cdots 2n-1}{n!}$
Extended Binomial Theorem
From the identity mentioned in the hint, one can get that the required sum is: $$S=\sum_{i=0}^{\infty} \frac{\binom{2i}{i}}{8^i}=\sum_{i=0}^{\infty} \left( \frac{-1}{2} \right) ^i \cdot \frac{-1}{2} \cdot \left( \frac{-1}{2}-1 \right) \cdot \left( \frac{-1}{2}-2 \right) \cdot \left( \frac{-1}{2}-3 \right) \cdots \left( \frac{-1}{2}-(i-1) \right) \cdot \frac{1}{i!}$$$$= \sum_{i=0}^{\infty} \binom{\frac{-1}{2}}{i} \left( \frac{-1}{2} \right) ^i =\left( 1+\left( \frac{-1}{2} \right) \right)^{\frac{-1}{2}}= \sqrt{2}$$
Honoured to be sniped by pco
Hexagrammum16
26.11.2019 18:31
If in place of $8^n$, we had $4^n$ (and the series starting from 1 instead of 0) then following the exact same routine we would obtain $S=(1+(-1))^{\frac{-1}{2}} -1$ which represents a divergent series. Now, using Stirling's Approximation on each of the individual terms, we see this new series is essentially $\frac{1}{\sqrt{\pi}} \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}}$. This shows that $\sum_{i=1}^{n} \frac{1}{\sqrt{i}}$ diverges.