Summation of Central Binomial Coefficients divided by even powers of $2$

Question

Whilst working out this problem the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$ and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.

Once the solution $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e. $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$

However, without knowing this a priori, how would we have approached this problem?

Answer 1

You may do the following to get rid of the powers of two and transform it into a standard sum: \begin{equation} \binom{2m}{m}=\frac{(2m)!}{m!m!}=\frac{2^m m! 1\cdot3\cdot5\cdot \cdots (2m-1)}{m!m!}=2^{2m} \frac{(1/2)(3/2)\cdots (m-1/2)}{m!}=2^{2m}\binom{m-1/2}{m}~~~~~~ (1) \end{equation} Then your sum becomes $$\sum_{m=0}^{n} \binom{m-1/2}{m}=\binom{n-1/2 +1}{n} =\binom{n+1-1/2}{n}=\frac{n+1}{n+1-1/2-n}\frac{n+1-1/2-n}{n+1}\binom{n+1-1/2}{n}=\frac{n+1}{1/2}\binom{n+1-1/2}{n+1}=2(n+1)\frac{1}{2^{2(n+1)}}\binom{2(n+1)}{n+1}=\frac{n+1}{2^{2{n+1}}}\binom{2n+2}{n+1},$$ where the first equality is just the usual identity $\sum_{j=0}^n \binom{r+j}{j} =\binom{r+n+1}{n}$ valid for any real $r$ and any non-negative integer $n$, and the last few steps were just trying to slightly transform the coefficient in a way that allowed application of $(1)$.

Answer 2

Another approach. By Euler/De Moivre's formula we have $$A_n=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{2n}(x)\,dx \tag{1}$$

since $\cos(x)^{2n} = \frac{1}{4^n}\sum_{j=0}^{2n}\binom{2n}{j}e^{(2n-j)ix}e^{-jix}$ and $\int_{-\pi}^{\pi}e^{kix}\,dx =2\pi \delta(k)$, hence only the contribute given by $j=n$ survives. It follows that $$ \color{red}{\sum_{n=0}^{N}\frac{1}{4^n}\binom{2n}{n}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-\cos^{2N+2}(x)}{\sin^2(x)}\,dx =(2N+2)\,A_{N+1}=\color{red}{\frac{N+1}{2^{2N+1}}\binom{2N+2}{N+1}}\tag{2}$$ by integration by parts ($\int\frac{dx}{\sin^2 x}=-\cot x$).

Answer 3

I've already accepted Sebastian Munoz's very good solution above.
This is a paraphrase of his solution for my own reference.

$$\begin{align} \sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m &\color{lightgrey}{=\sum_{m=0}^n\frac 1{2^{2m}}\frac {(2m)!}{m!m!}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac 1{2^{2m}}\frac{(2m)(2m-1)(2m-2)\cdots 3\cdot 2\cdot1}{(1\cdot 2\cdot3\cdot\cdots \cdot m)(1\cdot 2\cdot3\cdot\cdots \cdot m)}}\\ &\color{lightgrey}{=\sum_{m=0}^n \underbrace{\frac {(2m)(2m-2)(2m-4)\cdots 2}{2^m(1\cdot 2\cdot3\cdot\cdots \cdot m)}}_{=1}\cdot \frac{(2m-1)(2m-3)(2m-5)\cdots3\cdot 1}{2^{m}\cdot (1\cdot 2\cdot3\cdot\cdots \cdot m)}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac{\frac{2m-1}2\cdot \frac {2m-3}2\cdot \frac {2m-5}2\cdots \frac 32\cdot \frac 12}{1\cdot 2\cdot3\cdot\cdots \cdot m}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac{\left(m-\frac 12\right)\left(m-\frac 32\right)\left(m-\frac 52\right)\cdots \frac 32\cdot \frac 12}{1\cdot 2\cdot3\cdot\cdots \cdot m}}\\ &=\sum_{m=0}^n \binom {m-\frac 12}m\\ &=\binom {n+\frac 12}n\\ &\color{lightgrey}{=\frac {(n+\frac12)(n-\frac 12)(n-\frac 32)\cdots \frac 32}{1\cdot2 \cdot3 \cdot\quad\cdots \cdot n\qquad}}\\ &\color{lightgrey}{=\frac 1{2^n}\cdot \frac {(2n+1)(2n-1)(2n-3)\cdots 3}{1\cdot 2\cdot 3\cdot\cdots\cdot n}}\color{lightblue}{\cdot \frac{(2n+2)(2n)(2n-2)\cdots 2}{2^{n+1}(n+1)(n)(n-1)\cdots 1}\cdot \frac{n+1}{n+1}}\\ &=\frac {n+1}{2^{2n+1}}\cdot \frac{(2n+2)!}{(n+1)!(n+1)!}\\ &=\frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}\quad\blacksquare\end{align}$$

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Additional note: (added Oct 2018)

Note that $$\begin{align} \frac 1{\sqrt{1-4x}} &=(1-4x)^{-\frac 12}\\ &=\sum_{r=0}^\infty \binom {-\frac 12}r (-4x)^r = \sum_{r=0}^\infty \binom {-\frac 12}r (-1)^4 r^4x^r\tag{*} \\\ &=\sum_{r=0}^\infty \frac {\left(-\dfrac 12\right)_r}{r!} (-4x)^r\\ &=\sum_{r=0}^\infty \frac {(-1)^r(2r-1)!!}{(2^r)}\cdot \frac 1{r!}\cdot (-1)^r4^rx^r\\ &=\sum_{r=0}^\infty \frac {(2r-1)!!2^r}{r!}x^r\cdot \color{lightgrey}{\frac {r!}{r!}}\\ &=\sum_{r=0}^\infty \frac {(2r-1)!!(2r)!!}{r!r!}x^r\\ &=\sum_{r=0}^\infty\frac {(2r)!}{r!r!} x^r\\ &=\sum_{r=0}^\infty\binom {2r}r x^r\tag{**} \end{align}$$

From (*) and (**),

$$\begin{align} (-1)^r 4^r\binom {-\frac 12}r&=\qquad\;\binom {2r}r\\ \binom {-\frac 12}r&=\frac {(-1)^r}{4^r}\binom {2r}r\\ (-1)^r\binom {r-\frac 12}r&=\frac {(-1)^r}{4^r}\binom {2r}r\\ \binom {r-\frac 12}r&=\frac 1{\;2^{2r}}\;\binom {2r}r\end{align}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Indeed, we can use the result \begin{equation} \left.\sum_{n = 0}^{\infty}{2n \choose n}x^{n} \,\right\vert_{\ \verts{x}\ <\ 1/4} = {1 \over \root{1 - 4x}}\label{1}\tag{1} \end{equation} to evaluate the finite sum whenever we don't know the ' telescopic stuff '. It requires the use of a generating function:

\begin{align} \mc{F}\pars{z} & \equiv \sum_{n = 0}^{\infty}z^{n}\bracks{% \sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m}} \\[5mm] &\iff \sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m} = \bracks{z^{n}}\mc{F}\pars{z}\,, \quad\color{#f00}{\verts{z} < 1}\label{2}\tag{2} \end{align}

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\begin{align} \mc{F}\pars{z} & \equiv \sum_{n = 0}^{\infty}z^{n}\bracks{% \sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m}} = \sum_{m = 0}^{\infty}{1 \over 4^{m}}{2m \choose m} \sum_{n = m}^{\infty}z^{n} \\[5mm] & = \bracks{\sum_{m = 0}^{\infty}\pars{z \over 4}^{m}{2m \choose m}} \sum_{n = 0}^{\infty}z^{n} \\[5mm] & = {1 \over \root{1 - 4\pars{z/4}}}\,{1 \over 1 - z} = \pars{1 - z}^{-3/2} \\[5mm] & = \sum_{n = 0}^{\infty}{-3/2 \choose n} \pars{-1}^{n}z^{n} \qquad\pars{~\mbox{see expression}\ \eqref{1}~} \\[5mm] & = \sum_{n = 0}^{\infty}\bracks{{n + 1/2 \choose n}\pars{-1}^{n}}\pars{-1}^{n}z^{n} \end{align}

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\begin{equation} \mc{F}\pars{z} = \sum_{n = 0}^{\infty}{n + 1/2 \choose n}z^{n} \qquad\stackrel{\mrm{see\ expression}\ \eqref{2}}{\implies}\qquad \bbox[15px,#ffe,border:1px dashed navy]{\ds{% \sum_{m = 0}^{n}{1 \over 2^{2m}}{2m \choose m} = {n + 1/2 \choose n}}} \\ \end{equation}

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The last expression can be rewritten as the OP reported answer $\ds{{n + 1 \over 2^{2n + 1}}{2n + 2 \choose n + 1}}$ by means of the $\ds{\Gamma}$-Duplication Formula.

Answer 5

Using an Extension of Pascal's Rule $$ \begin{align} \sum_{m=0}^n\frac1{2^{2m}}\binom{2m}{m} &=\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\tag{1a}\\ &=\sum_{m=0}^n\binom{m-\frac12}{m}\tag{1b}\\ &=\sum_{m=0}^n\left[\binom{m+\frac12}{m}-\binom{m-1+\frac12}{m-1}\right]\tag{1c}\\ &=\binom{n+\frac12}{n}\tag{1d}\\[6pt] &=\frac{(2n+1)!!}{(2n)!!}\tag{1e}\\[6pt] &=\frac{2n+1}{2^{2n}}\binom{2n}{n}\tag{1f} \end{align} $$ Explanation:
$\text{(1a)}$: $\frac{\color{#C00}{(2m)!}}{\color{#C00}{2^mm!}\,\color{#090}{2^mm!}}=\frac{\color{#C00}{(2m-1)!!}}{\color{#090}{(2m)!!}}$
$\text{(1b)}$: divide numerator and denominator by $2^m$
$\text{(1c)}$: apply $(4)$ with $\alpha=\frac12$
$\text{(1d)}$: telescoping sum
$\text{(1e)}$: multiply numerator and denominator by $2^n$
$\text{(1f)}$: $\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)\color{#C00}{(2n-1)!!}}{\color{#090}{(2n)!!}}=\frac{(2n+1)\color{#C00}{(2n)!}}{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}$

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Extension of Pascal's Rule

Newton's Generalized Binomial Theorem says $$ \begin{align} \sum_{m=0}^\infty\binom{m+\alpha}{m}x^m &=\sum_{m=0}^\infty(-1)^m\binom{-1-\alpha}{m}x^m\tag{2a}\\ &=(1-x)^{-1-\alpha}\tag{2b} \end{align} $$ Explanation:
$\text{2a}$: convert to negative binomial coefficient
$\text{2b}$: Binomial Theorem

Thus, $$ \begin{align} \sum_{m=0}^\infty\binom{m-1+\alpha}{m}x^m &=(1-x)^{-\alpha}\tag{3a}\\[6pt] &=(1-x)(1-x)^{-1-\alpha}\tag{3b}\\[9pt] &=(1-x)\sum_{m=0}^\infty\binom{m+\alpha}{m}x^m\tag{3c}\\ &=\sum_{m=0}^\infty\binom{m+\alpha}{m}\left(x^m-x^{m+1}\right)\tag{3d}\\[3pt] &=\sum_{m=0}^\infty\left[\binom{m+\alpha}{m}-\binom{m-1+\alpha}{m-1}\right]x^m\tag{3e} \end{align} $$ Explanation:
$\text{(3a)}$: apply $\text{(2b)}$
$\text{(3b)}$: factor out $(1-x)$
$\text{(3c)}$: apply $\text{(2b)}$
$\text{(3d)}$: distribute $(1-x)$
$\text{(3e)}$: substitute $m\mapsto m-1$ in the subtrahend

Thus, for arbitrary $\alpha\in\mathbb{R}$, we can extend Pascal's Rule to $$ \binom{m-1+\alpha}{m}+\binom{m-1+\alpha}{m-1}=\binom{m+\alpha}{m}\tag4 $$