Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$

Question

Mathematica tells me that

$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$

Although I have not been able to come up with a proof.

Proofs, hints, or references are all welcome.

Answer 1

You could consider the integral $$ \int_{0}^{1} (1-x^2)^n dx .$$

Answer 2

Sums of the form $$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.

Answer 3

I can't offer specific help, but I'd recommend thumbing through Concrete Mathematics looking for techniques. It has many sums that look similar, though of course the difficulty is in the details.

There's also the book A=B, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.

Answer 4

Your identity is a special case of the Chu-Vandermonde identity.

${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$

with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.

Answer 5

There are a powerful algorithms generalizing telescopy (Gosper, Zeilberger et al.) that easily tackle this case of the Chu-Vandermonde identity and much more complicated sums. For example, see this paper which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of Euler's pentagonal number theorem is simply the dual of a special case of a q-Chu-Vandermonde.

Answer 6

$$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$

We have:

$$\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\ &=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\&+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]\end{align}$$ Therefore, $$S_n=\dfrac{2n}{2n+1}S_{n-1}+0 \Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2} ... \Rightarrow S_1=\dfrac{2}{3}S_0$$ and $S_0=1$

Hence, $$S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$$

Answer 7

In trying to prove

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

we introduce

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

This has the property that for $0\le k\le n$

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

Note that the residue of $f(z)$ at infinity is zero by inspection and residues sum to zero so that we have

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

This is the claim.