How find $\lim_{n\to+\infty}\sum_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}}$? How find this limit $\displaystyle\lim_{n\to+\infty}\sum_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}}$
 A: We first notice that $\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}}$ vanishes when $n$ is odd. So we need only consider the case when $n$ is even. 
$\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}} = \sum\limits_{k=0}^{n}(-1)^{k}\left(\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma{(n-k+1)}}\right)^{1/2}$
Consider, the function $\displaystyle f(z) = \left(\frac{\Gamma(n+1)}{\Gamma(z+1)\Gamma(n-z+1)}\right)^{1/2}\csc \pi z$
Hence, $f(z)$ has simple poles at $z = 0,1,\cdots,n$ with residue $\dfrac{(-1)^k}{\pi}$ at these points for $k=0,1,\cdots,n$.
Integrating $f(z)$ over rectangular contour with four vertices $(-1/2\pm iT),(n+\frac{1}{2}\pm iT)$ call it $C_T$,
By the residue theorem,
$$\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}} = \frac{1}{2i}\int\limits_{C_T} f(z)\,dz$$
Using the reflection formula $\displaystyle \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}$ we have:
$$f(z)^2 = \frac{\Gamma(n+1)\Gamma(1-z)}{\pi z \Gamma(n+1-z)}\csc \pi z = \frac{\Gamma (n+1)\csc \pi z}{\pi z (n-z)(n-1-z)\cdots (1-z)}$$
We estimate the integral on the horizontal lines first,
When $z = x\pm iT$, for a large enough $T$ such that $|z - k| \ge T$ for all $k =0,1,\cdots,n$
$$|f(z)|^2 < \frac{\Gamma(n+1)}{\pi T^{n+1}}. \frac{2}{|e^{i(x+iT)} - e^{-i(x+iT)}|} \sim \frac{\Gamma(n+1)e^{-\pi T}}{\pi T^{n+1}}$$
Hence, $|f(z)|^2 \to 0$ as $T \to \infty$.
Thus, $$\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}} = \frac{1}{2i}\int\limits_{n+\frac{1}{2}-i\infty}^{n+\frac{1}{2}+i\infty} f(z)\,dz - \frac{1}{2i}\int\limits_{-\frac{1}{2}-i\infty}^{-\frac{1}{2}+i\infty} f(z)\,dz$$
Putting $\displaystyle z = n+\frac{1}{2}+it$, in the first integral and $\displaystyle z = -\frac{1}{2}+it$, in the second integral, and since $n$ is even, 
$$\displaystyle F(n) = \sum\limits_{k=0}^{n}(-1)^{k}\sqrt{\binom{n}{k}} = \int_{-\infty}^{\infty} \left(\frac{\Gamma(n+1)\Gamma\left(\frac{1}{2}+it\right)}{\pi\Gamma\left(n+\frac{3}{2}+it\right)\cosh \pi t}\right)^{1/2}\,dt$$
Since, $\displaystyle \left|\frac{\Gamma\left(\frac{1}{2}+it\right)}{\Gamma\left(n+\frac{3}{2}+it\right)}\right| < \frac{1}{(n+\frac{1}{2})(n -\frac{1}{2}) \cdots (\frac{1}{2})} = \frac{\pi}{\Gamma (n+\frac{3}{2})}$
We have $$|F(n)| \le \left(\frac{\Gamma(n+1)}{\sqrt{\pi}\Gamma(n+\frac{3}{2})}\right)^{1/2}\int_{-\infty}^{\infty} \frac{dt}{\sqrt{\cosh \pi t}}$$
Hence, $$\lim\limits_{n \to \infty} F(n) = 0$$
