The summation $\sum_{k=0}^{2n} (-1)^k \frac{{4n \choose 2k}}{{2n \choose k}}=\frac{1}{1-2n}$

Question

This summation has been created by perseverance over a long period of time by using the results from the Table of Series and Integrals By I.S. Gradshteyn and I.M. Rhyzik [GR].

The idea was to make a summation of binomial coefficients by utilizing the Integral representation of the reciprocal of the Beta (Integral/function): $$B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)\tag{1}$$ as
$$\frac{1}{B(x,y)}=\frac{(x+y-1)}{\pi} \int_{0}^{\pi/2}
\cos[(x-y)t]~\cos^{x+y-1}t ~dt~ \mbox{[GR, p. 959]}.\tag{2}$$
Let $$S_n=\sum_{k=0}^{2n} (-1)^k \frac{{4n \choose 2k}}{{2n \choose k}}. \tag{3}$$
We change the summand into Gamma function using $\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}} \Gamma(z+1/2)\Gamma(z)$, we
get $$S_n=\sqrt{\pi} \sum_{k=0}^{2n} (-1)^k \frac{\Gamma(z)}{\Gamma(z-k) \Gamma(k+1/2)},~~ z=2n+1/2.$$
We can re-write $S_n$ using (1) as
$$ S_n=\sqrt{\pi} \frac{\Gamma(z)}{\Gamma(z+1/2)} \sum_{k=0}^{n} \frac{(-1)^k}{B(z-k,k+1/2)}.$$
Then $$S_n= \frac{2^{2n+1}}{\sqrt{\pi} \Gamma(2n)} \int_{0}^{\pi/2} \sum_{k=0}^{n} (-1)^k \cos[(k-n)t] \cos^{2n-1} t ~dt.$$
Note that [GR, p. 37]
$$
\sum_{k=0}^{2n} (-1)^k \cos [(k-n)t] =\sum_{m=-n}^{n} (-1)^{m+n} \cos 2m t= (-1)^n \left[1+2\sum_{m=1}^{n}(-1)^m \cos 2mt\right]=\frac{\cos(2n+1)t}{\cos t}.
$$
We obtain
$$
S_n=\frac{2^{2n}\Gamma(2n+1/2)}{\sqrt{\pi}\Gamma(2n)} \int_{0}^{\pi/2} \cos(2n+1)t ~ \cos^{2n-2}t ~dt.
$$
Using the value of this integral from [GR, p. 416], we can write
$$S_n=\frac{2^{2n}\Gamma(2n+1/2)}{\sqrt{\pi}\Gamma(2n)}\left(\frac{\pi} {2^{2n-1}(2n-1) B(2n+1/2,-1/2)}\right).$$
Finally by opening $B(x,y)$ and using $\Gamma(-1/2)=-2 \sqrt{\pi},$ we get
$$ S_n=\frac{1}{1-2n}.$$
This sum (3) can be checked to be correct and consistent but it came mostly by the Table! So the question is: Can one give an alternate proof of (3) by hand which is simpler, more direct and characteristic?

Answer 1

By using W-Z method, we show a more general identity: for $n\geq 2$, $$\sum_{k=0}^{n} (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}=\frac{1+(-1)^n}{2(1-n)}.$$

Let $$F(n,k)= (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}(n-1).$$ Then for $k=0,1,\dots,n$, $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ with $$G(n,k)=(-1)^k\frac{\binom{2n}{2k-2}}{\binom{n+1}{k}}(n+1).$$ Hence $$\begin{align} \sum_{k=0}^{n+1}F(n+1,k)-\sum_{k=0}^{n}F(n,k)&=F(n+1,n+1)+ \sum_{k=0}^{n}(G(n,k+1)-G(n,k))\\ &=F(n+1,n+1)+G(n,n+1)-G(n,0)\\ &=(-1)^{n+1}n+(-1)^n(n+1)-0\\ &=(-1)^n. \end{align}.$$ Finally, for $n\geq 2$, $$\sum_{k=0}^{n} (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}=\frac{1}{n-1}\sum_{k=0}^{n}F(n,k)=\frac{1}{n-1}\sum_{k=2}^{n}(-1)^{k-1}=\frac{1+(-1)^n}{2(1-n)}.$$