$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$?

Question

How can you compute the following summation:

$$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$$

Answer 1

The binom formula gives that for all $\alpha \in \mathbb{C}$, $$\sum\limits_{k\geqslant 0} \binom{\alpha}{k}x^k = (1+x)^\alpha$$

By taking $\alpha = -1/2$, we get $$\sum\limits_{k\geqslant 0} \binom{-1/2}{k} x^k = (1+x)^{-1/2}$$

Then we remark that $\binom{-1/2}{k}=(-1)^k \binom{k+1/2}{k}$.

So we have $$\sum\limits_{k\geqslant 0} \binom{k+1/2}{k}(-1)^k x^k = \frac{1}{\sqrt{1+x}}$$.

And as $\binom{k+1/2}{k}=\binom{2k}{k}\cdot \frac{1}{4^k}$, we get the answer $$\sum\limits_{k\geqslant 0} \binom{2k}{k}(-1)^k \frac{x^k}{4^k} = \frac{1}{\sqrt{1+x}}$$

By replacing with $x=1$, we get $\frac{1}{\sqrt{2}}$.

Answer 2

It probably was $$ \sum_{k\geq 1}\frac{\binom{2k}{k}}{(2k-1)4^k} = 1$$ or the equivalent (via $k\mapsto k+1$) $$ \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(k+1)} = 2$$ which are given by the Maclaurin series of $\sqrt{1-x}$ or by creative telescoping. Since $$ \int_{0}^{\pi/2}(\cos\theta)^{2k}\,d\theta = \frac{\pi}{2\cdot 4^k}\binom{2k}{k} $$ holds by integration by parts, we also have $$ \sum_{k\geq 0}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k\geq 0}(-1)^k(\cos\theta)^{2k}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta} $$ and via $\theta=\arctan t$ $$ \sum_{k\geq 0}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{2+t^2} =\frac{1}{\sqrt{2}}.$$