# Proving $\sum\limits_{k=0}^{n-1}\frac{\left(-1\right)^{n-k-1}\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{2^{2n}\binom{n}{k}} = (-1)^{n-1}$

I found the below combinatorial identity seems true:

$$\sum\limits_{k=0}^{n-1}\frac{\left(-1\right)^{n-k-1}\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{2^{2n}\binom{n}{k}} = (-1)^{n-1}$$

but I have no idea how to prove it, anyone has some ideas?

Here is where this problem comes from.

I want to prove this Fourier series expansion

$$\log\left(2\cos\frac{\theta}{2}\right)=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}\cos n\theta}{n}$$

then I try to calculate the coefficients by evaluating the following integral

$$\begin{eqnarray} a_{n}&=&\frac{2}{\pi}\int_{0}^{\pi}\cos\left(n\theta\right)\log\left(2\cos\frac{\theta}{2}\right)d\theta\\&=&\frac{2}{n\pi}\int_{0}^{\pi}\frac{\sin\left(n\theta\right)\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}d\theta\\&=&\frac{\left(-1\right)^{n-1}}{n}\\A_{n}&=&\int_{0}^{\pi}\sin\left(n\theta\right)\tan\frac{\theta}{2}d\theta\\&=&2\int_{0}^{\frac{\pi}{2}}\sin\left(2n\theta\right)\tan\theta d\theta\\&=&2\sum_{k=0}^{n-1}\left(-1\right)^{n-k-1}\left(\begin{array}{c} 2n\\ 2k+1 \end{array}\right)\int_{0}^{\frac{\pi}{2}}\left(\cos\theta\right)^{2k}\left(\sin\theta\right)^{2n-2k}d\theta\\&=&\sum_{k=0}^{n-1}\left(-1\right)^{n-k-1}\left(\begin{array}{c} 2n\\ 2k+1 \end{array}\right)\frac{\Gamma\left(\frac{2k+1}{2}\right)\Gamma\left(\frac{2n-2k+1}{2}\right)}{\Gamma(n+1)}\\&=&\pi\sum_{k=0}^{n-1}\left(-1\right)^{n-k-1}\left(\begin{array}{c} 2n\\ 2k+1 \end{array}\right)\frac{\left(\begin{array}{c} 2k\\ k \end{array}\right)k!\left(\begin{array}{c} 2n-2k\\ n-k \end{array}\right)(n-k)!}{2^{2n}n!}\\&=&\pi\sum_{k=0}^{n-1}\frac{\left(-1\right)^{n-k-1}\left(\begin{array}{c} 2n\\ 2k+1 \end{array}\right)\left(\begin{array}{c} 2k\\ k \end{array}\right)\left(\begin{array}{c} 2n-2k\\ n-k \end{array}\right)}{2^{2n}\left(\begin{array}{c} n\\ k \end{array}\right)}\\&=&2\pi\left(-1\right)^{n-1}? \end{eqnarray}$$

therefore, I have two questions:

(1) is there a direct way to evaluate the integral $$A_n$$ ?

(2) how to prove the identity in a combinatoric way ?

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Jul 26, 2023 at 10:05
• Please edit to include your efforts. It appears to be an interesting result...I think you'd get a better reception if you at least indicated how far you have confirmed it. Context would also help...where did this summation arise?
– lulu
Commented Jul 26, 2023 at 10:39
• Some simplification would also help. Note that multiplicative factors of $(-1)^{n-1}$ appear on both sides and could simply be cancelled. Also, the various binomial symbols certainly allow for some cancellation (not sure if that helps or not but it is worth a look).
– lulu
Commented Jul 26, 2023 at 10:40
• Note that $$\frac{\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{\binom{n}{k}}=\frac{2n-2k}{2k+1}\binom{2n}{n}\binom{n}{k}.$$ Canceling $(-1)^{n-1}$ on both sides and clearing constants in $k$, we see that we only need to prove that $$\sum_{k=0}^{n-1}(-1)^k \frac{2n-2k}{2k+1}\binom{n}{k}=\frac{2^{2n}}{\binom{2n}{n}}.$$ For $n>0$, $\sum_{k=0}^{n-1}(-1)^k \binom{n}{k}=0.$ Add that to the previous equation to obtain $$\sum_{k=0}^{n-1}(-1)^k\frac{2n+1}{2k+1}\binom{n}{k}=\frac{2^{2n}}{\binom{2n}{n}},$$ i.e. $$\sum_{k=0}^{n-1}(-1)^k \frac{1}{2k+1}\binom{n}{k}=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}.$$ Commented Jul 26, 2023 at 14:36
• quite an inspired transformation Commented Jul 27, 2023 at 11:58

The claim is equivalent to $$\sum\limits_{k=0}^{n-1}\frac{\left(-1\right)^{k}\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{4^n\binom{n}{k}} = 1$$

Brute force on this seems to work, but I don't have a very neat proof. The only non-standard identity we need (non-standard to me, anyway) is

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

To prove this, note that $$\left(1-x^2\right)^{n-1}=\sum_{k=0}^{n-1} \binom{n-1}{k}(-1)^k x^{2k}$$

so that $$\int_0^1 \left(1-x^2\right)^{n-1} dx =\sum_{k=0}^{n-1} \frac{(-1)^k}{2k+1} \binom{n-1}{k}$$

Substituting $$x=\sin{u}$$, this integral becomes $$\int_0^\frac{\pi}{2} \cos^{2n-1}{(u)} du =:I(2n-1)$$

and it's a standard result using integration by parts that $$I(N)=\frac{N-1}{N} I(N-2)$$ where $$I(0)=\frac{\pi}{2}$$ and $$I(1)=1$$.

The form we need is that $$I(2n-1) = \frac{2n-2}{2n-1} I(2n-3)$$

and it's easy to check by induction that this is indeed $$I(2n-1)=\frac{4^n}{2n\binom{2n}{n}}$$

With this, the rest is just manipulating binomial coefficients.

\begin{align} \sum\limits_{k=0}^{n-1}\frac{\left(-1\right)^{k}\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{4^n\binom{n}{k}} &= \sum\limits_{k=0}^{n-1}\frac{\left(-1\right)^{k}(2n)! (2k)! (2n-2k)! k!(n-k)! }{4^n n! (2k+1)!(2n-2k-1)!k!^2 (n-k)!^2} \\ &= \sum\limits_{k=0}^{n-1}\frac{2\left(-1\right)^{k}(2n)! }{4^n n! (2k+1) k! (n-k-1)!} \\ &= \sum\limits_{k=0}^{n-1} \frac{2(2n)!}{4^n n!} \frac{(-1)^k}{(2k+1) k! (n-k-1)!} \\ &= \sum\limits_{k=0}^{n-1} \frac{2n(2n)!}{4^n (n!)^2} \frac{(-1)^k (n-1)!}{(2k+1) k! (n-k-1)!} \\ &= \frac{2n}{4^n}\binom{2n}{n} \sum\limits_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\binom{n-1}{k} \\ &=1 \end{align}

by the result above, so the claim is proved.

• Apologies for any typos - the LaTeX gets horrendous with expressions like these. Any thoughts on making this clearer would be appreciated (although I strongly suspect there's a better way). Commented Jul 26, 2023 at 12:44
• Thanks for your effort. I have provided some more background on this problem, I hope this will be helpful. Commented Jul 27, 2023 at 12:00
• @ChrisLewis nice answer +1 Commented Aug 2 at 18:53

Now that this question is no longer closed, I can post my previous comment as an answer.

Note that $$\frac{\binom{2n}{2k+1}\binom{2k}{k}\binom{2n-2k}{n-k}}{2^{2n}\binom{n}{k}}=\frac{2n-2k}{2k+1}\binom{n}{k}\binom{2n}{n}.$$ This holds for $$k=n$$ as well, we just get $$0=0$$. Canceling $$(−1)^{n-1}$$ on both sides and clearing constants in $$k$$, we see that we only need to prove that $$\sum_{k=0}^{n}(-1)^k\frac{2n-2k}{2k+1}\binom{n}{k}=\frac{2^{2n}}{\binom{2n}{n}}.$$ Now $$n\ge 1$$, so $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}=0$$. Adding that to the previous equation, we obtain $$\sum_{k=0}^{n}(-1)^k\frac{2n+1}{2k+1}\binom{n}{k}=\frac{2^{2n}}{\binom{2n}{n}},$$ i.e. $$\sum_{k=0}^{n}(-1)^k\frac{1}{2k+1}\binom{n}{k}=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}=\frac{(2n)!!}{(2n+1)!!}.$$ This formula has been proven and re-proven here many times, just use Approach0 to search for it. The basic idea of one of the solutions is that $$\sum_{k=0}^{n}(-1)^k\frac{1}{2k+1}\binom{n}{k} =\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-1)^kx^{2k}\,dx =\int_{0}^{1}\left(1-x^2\right)^n\,dx,$$ and integration by parts yields, for $$n\ge 1$$, $$\begin{split} I_n&=\int_{0}^{1}\left(1-x^2\right)^n\,dx=\left[x\left(1-x^2\right)^n\right]_{0}^{1}-\int_{0}^{1}x(-2x)n\left(1-x^2\right)^{n-1}\,dx\\ &=2n\int_{0}^{1} x^2\left(1-x^2\right)^{n-1}\,dx=2n(I_n-I_{n-1}), \end{split}$$ so $$I_n=\frac{2n}{2n+1}I_{n-1}, \qquad n\ge 1,$$ and $$I_0=1$$, which is easy to see.

• Thanks for your effort, Alex. Commented Jul 31, 2023 at 6:42