Proof that $\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$

Question

I saw in this paper the following identity: $$\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$$ I have a pervious post on an integral quite closely related to this identity, but I still do not know how to derive/prove the identity. I'm really not that good at combinatorics or evaluating series, so I don't know how to start, which is why I don't have any attempts to show you. Please explain your steps thoroughly.

Edit: I know there other posts on this series, but I did not get from them the proof I was satisfied by.

Answer 1

$$f(x)=(1-x^2)^n=\sum_{k=0}^n \binom{n}{k}x^{2k}(-1)^k$$

$$F(n)=\int_0^1(1-x^2)^n dx=\sum_{k=0}^n \frac{(-1)^k}{2k+1} \binom nk\tag{1}$$

On the other side (see proof here):

$$F(n)=\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$$

It's a simple exercise to show that:

$${(2n)!!\over (2n+1)!!}=\frac{4^n}{(2n+1){2n\choose n}}$$

...which completes the proof.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} = {4^{n} \over \pars{2n + 1}{2n\choose n}}:\ {\LARGE ?}}$.

\begin{align} \sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} & = \sum_{k = 0}^{n}\pars{-1}^{k}\pars{\int_{0}^{1}t^{2k}\dd t}{n \choose k} = \int_{0}^{1}\bracks{\sum_{k = 0}^{n}{n \choose k}\pars{-t^{2}}^{k}}\dd t \\[5mm] & = \int_{0}^{1}\pars{1 - t^{2}}^{n}\dd t \\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, {1 \over 2}\ \overbrace{\int_{0}^{1}t^{-1/2}\,\pars{1 - t}^{n}\,\dd t} ^{\ds{\mrm{B}\pars{1/2,n + 1}}}\qquad\pars{~\mrm{B}:\ Beta\ Function~} \\[5mm] & = {1 \over 2}\,{\Gamma\pars{1/2}\Gamma\pars{n + 1} \over \Gamma\pars{n + 3/2}}\qquad\pars{~\Gamma:\ Gamma\ Function~} \\[5mm] & = {1 \over 2}\,{\root{\pi}\Gamma\pars{n + 1} \over \root{2\pi}2^{-3/2 - 2n}\Gamma\pars{2n + 2}/ \Gamma\pars{n + 1}}\label{1}\tag{1} \end{align} In the last expression, I used $\ds{\Gamma\pars{1/2} = \root{\pi}}$ and the Gamma Duplication Formula.

\eqref{1} becomes:

\begin{align} \sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} & = {4^{n}\pars{n!}^{2} \over \pars{2 n + 1}!} = {4^{n} \over \pars{2n + 1}\bracks{\pars{2n}!/\pars{n!}^{2}}} = \bbx{{4^{n} \over \pars{2n + 1}{2n \choose n}}} \end{align}

Answer 3

The Binomial Theorem says $$ \sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}=\left(1-x^2\right)^n $$ Integrating both sides over $[0,1]$ gives $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k} &=\int_0^1\left(1-x^2\right)^n\,\mathrm{d}x\\ &=\frac12\int_0^1\left(1-x\right)^nx^{-1/2}\,\mathrm{d}x\\ &=\frac12\frac{\Gamma(n+1)\,\Gamma(1/2)}{\Gamma(n+3/2)}\\ &=\frac12\frac{n!}{\frac12\frac32\cdots\left(n+\frac12\right)}\\ &=\frac{2^nn!}{1\cdot3\cdots(2n+1)}\\ &=\frac{2^nn!2^nn!}{(2n+1)!}\\ &=\frac{4^n}{(2n+1)\binom{2n}{n}} \end{align} $$ Using the Beta Function.

Answer 4

I was actually thinking about exactly this sum a few days ago, and found a few ways to evaluate it. I'm on my phone right now so I can't tpe it up all pretty, but I'll give the general idea.

Method 1

(-1)^k / (2k+1) = 1/i * integral from 0 to i of x^2k dx

Multiplying by binomial coefficients gives you that the sum you're looking for is

integral from 0 to i of (1+x^2)^n dx

A simple integration by parts proves your formula via induction.

Method 2

1 / (k + 1/2) = integral from 0 to 1 of x^(k-1/2) dx

Multiplying by binomial coefficients and summing you get that the sum is equal to half of

integral from 0 to 1 of (1-x)^n * x^(-1/2) dx

But that's the beta function, and if you evaluate it using the formula involving the gamma function you get the correct answer.

Notice that method 2 easily generalizes to the case where we replace 2k+1 with xk+y where x and y are arbitrary real numbers, and indeed to the case where n is any real number.