# Proving a binomial sum identity

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.

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

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Aha this is the integral I was searching for! Using the integral $\int_0^1 (1-t)^n \, dt$ I was able to prove $\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{k+1} = \frac{1}{n+1}$ I thought there should be a corresponding integral to my problem and after seeing your answer the integral I was searching for is obvious now! – yjj Sep 8 '10 at 11:02

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.

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

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

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As stated, you divide by zero since you sum to $k=n$ and $n-k$ is in the denominator. – Julian Kuelshammer Nov 29 '12 at 7:36
Wah! Thanks for pointing out errors in my solution. – hxthanh Dec 1 '12 at 12:12
Fix this problem as follows sum for $k=0$ to $n-1$ and $k=n$ is calculated separately – hxthanh Dec 1 '12 at 12:34
You can edit your post accordingly. – Julian Kuelshammer Dec 1 '12 at 12:37

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.

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It's on page 181. – J. M. Sep 8 '10 at 11:05

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.

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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.

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