Proving the divergence of $\sum {2j \choose j} (\frac14)^j$

Question

I need to evaluate the series:$$\sum_{j=0}^\infty {2j \choose j} \left(\frac14\right)^j$$ Wolfram Alpha claims that it is divergent by the comparison test. The partial sums graph also seems to confirm this -- however, I have no idea what I can compare the sum to.

Answer 1

You can compare to the harmonic series $\sum_j 1/j$. Namely if $T(j) = {2 j \choose j} 4^{-j}$, $$ \dfrac{(j+1) T(j+1)}{j T(j)} = \frac{j+1}{4 j} \frac{(2j+2)!}{(j+1)!^2} \frac{j!^2}{(2j)!} = \frac{(j+1)(2j+2)(2j+1)}{4 j(j+1)^2} = \frac{ 2j+1}{2j} > 1$$ so that $$T(j) > \frac{T(1)}{j}$$

Answer 2

Look at row $2n$ in the Pascal triangle. The sum of all $2n+1$ terms is $2^{2n}= 4^n$ . Since the central binomial coefficient is the largest number in that row, we have $4^n \le (2n+1){{2n} \choose n}$. Hence $$ \frac{1}{4^n}{{2n} \choose n} \ge \frac{1}{2n+1} \ge \frac{1}{2}\frac{1}{n+1} $$ and the divergence follows from comparison with the harmonic series.

Answer 3

Assuming $n\geq 2$ we have: $$\begin{eqnarray*}\frac{1}{4^n}\binom{2n}{n}=\frac{(2n)!}{4^n n!^2}&=&\frac{(2n-1)!!}{(2n)!!}\\&=&\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\\&=&\frac{1}{2}\sqrt{\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2}\\&=&\frac{1}{2}\sqrt{\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1}}\\&\geq&\frac{1}{2}\sqrt{\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(2k-1)^2}}\\&\geq&\frac{1}{2}\sqrt{\frac{1}{n}\left(\frac{\pi^2}{8}-\frac{1}{4n}\right)}\\&\geq &\frac{\pi}{4\sqrt{2n}}\left(1-\frac{1}{9n}\right)\end{eqnarray*}$$ and a similar inequality holds in the opposite direction, too. It follows that $$ K_1\sqrt{N}\leq \sum_{n=1}^{N}\frac{1}{4^n}\binom{2n}{n} \leq K_2 \sqrt{N} $$ for two positive constants $K_1,K_2$. Actually: $$ \sum_{n=0}^{N}\frac{1}{4^n}\binom{2n}{n} = \frac{N+1}{2\cdot 4^N}\binom{2N+2}{N+1}\sim\frac{2}{\sqrt{\pi}}\sqrt{N}. $$

Answer 4

You can use Stirling's formula to find an asymptotic for $\binom{2j}j$.

Alternatively use $$\frac1{\sqrt{1-x}}=\sum_{j=0}^\infty\binom{2j}j\frac{x^j}{4^j}.$$

Answer 5

HINT: for the finite sum is hold $$ \sum_{j=0}^m \binom{2j}{j}=2\,{\frac { \left( m+1 \right) \left( 2\,m+2 \right) !\, \left( 1/4 \right) ^{m+1}}{ \left( \left( m+1 \right) ! \right) ^{2}}} $$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Indeed, $\bbox[10px,#ffe,border:1px dotted navy]{\ds{{2j \choose j} = {-1/2 \choose j}\pars{-4}^{\,j}}}$ such that \begin{align} \sum_{j = 0}^{\infty}{2j \choose j}z^{\,j} & = \sum_{j = 0}^{\infty}{-1/2 \choose j}\pars{-4z}^{\,j} \,\,\,\stackrel{\verts{-4z}\ <\ 1}{=}\,\,\, \bracks{\vphantom{\large A}1 + \pars{-4z}}^{\,-1/2} = \bbx{{1 \over \root{1 - 4z}}\,,\quad\verts{z} < {1 \over 4}} \end{align}