A finite sum closely related to the binomial series

Question

After integrating a certain function in terms of the binomial theorem I was left with a finite sum that I haven't made any progress on yet.

$$ \sum_{k=0}^{2n} \frac{(2n)!}{k!(2n-k)!} \frac{(-1)^k}{2k+1} $$

The limit is always an even number. Any help summing this would be greatly appreciated.

Answer 1

Here is another variation based upon the Beta function.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^{2n}}\color{blue}{\frac{(2n)!}{k!(2n-k)!} \frac{(-1)^k}{2k+1}} &=\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k\int_{0}^1x^{2k}\,dz\tag{1}\\ &=\int_{0}^1\sum_{k=0}^{2n}\binom{2n}{k}\left(-z^2\right)^k\,dz\\ &=\int_{0}^1\left(1-z^2\right)^{2n}\,dz\\ &=\frac{1}{2}\int_{-1}^1\left(1-z^2\right)^{2n}\,dz\tag{2}\\ &=2^{4n-1}\int_{-1}^1\left(\frac{1+z}{2}\right)^{2n}\left(1-\frac{1+z}{2}\right)^{2n}\,dz\tag{3}\\ &=2^{4n}\int_{0}^1u^{2n}(1-u)^{2n}\,du\tag{4}\\ &\,\,\color{blue}{=\frac{2^{4n}}{4n+1}\binom{4n}{2n}^{-1}}\tag{5} \end{align*}

Note that we have $\frac{2^{4n}}{4n+1}\binom{4n}{2n}^{-1}=\frac{2^{4n+1}}{2n+1}\binom{4n+2}{2n+1}^{-1}$ in accordance with @MarkoRiedel's answer.

Comment:

• In (1) we use $\frac{1}{k+1}=\int_{0}^1z^k\,dz$.

• In (2) we use that the even function $f(z)=(1-z^2)^{2n}$ is symmetric with respect to the $y$-axis.

• In (3) we use $1-z^2=(1+z)(1-z)=\frac{1}{4}\left(\frac{1+z}{2}\right)\left(1-\frac{1+z}{2}\right)$.

• In (4) we substitute $u=\frac{1+z}{2}, du=\frac{1}{2}dz$.

• In (5) we we write the reciprocal of a binomial coefficient using the beta function \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz \end{align*}

Answer 2

In terms of the Gamma function, your sum can be written as

$$ \dfrac{(2n)!\; \Gamma(3/2)}{\Gamma(2n+3/2)}$$ (the $\sqrt{\pi}/2$ in zjs's comment is $\Gamma(3/2)$).

Answer 3

The basic technique which has appeared several times at MSE is to introduce the function

$$f(z) = \frac{(2n)!}{2z+1} \prod_{q=0}^{2n} \frac{1}{z-q}$$

which has the property that for $0 \le k\le 2n$

$$\mathrm{Res}_{z=k} f(z) = \frac{(2n)!}{2k+1} \prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^{2n} \frac{1}{k-q} \\ = \frac{(2n)!}{2k+1} \frac{1}{k!} \frac{(-1)^{2n-k}}{(2n-k)!} = {2n\choose k} \frac{(-1)^k}{2k+1}.$$

With residues summing to zero and the residue at infinity being zero by inspection we thus have

$$\sum_{k=0}^{2n} {2n\choose k} \frac{(-1)^k}{2k+1} = - \mathrm{Res}_{z=-1/2} f(z).$$

This is

$$- \frac{1}{2} (2n)! \prod_{q=0}^{2n} \frac{1}{-1/2 - q} = -\frac{1}{2} (2n)! 2^{2n+1} (-1)^{2n+1} \prod_{q=0}^{2n} \frac{1}{2q+1} \\ = (2n)! 2^{2n} \frac{(2n+1)! 2^{2n+1}}{(4n+2)!}.$$

We get at last

$$\bbox[5px,border:2px solid #00A000]{ \frac{2^{4n+1}}{2n+1} {4n+2\choose 2n+1}^{-1} .}$$