Limit of sum with binomial coefficient Prove that: 
$$\lim_{n\to +\infty}\sum_{k=0}^{n}(-1)^k\sqrt{{n\choose k}}=0$$
I completely don't know how to approach. Is it very difficult?
 A: First note that $f(z)=\frac{\sin \pi z}{\pi z(1-z)(1-\frac{z}{2})\cdots(1-\frac{z}{n})}$ satisfies $f(z)=\binom{n}{k}$ for any integer $k$.
Because $f(z)$ has no zeros in $-1<Re(z)<n+1$,$\sqrt{f(z)}$ (taken as positive at the origin) is analytic there.  By the Residue Theorem, we have
$$\sum_{k=0}^{n}(-1)^{k} \sqrt{\binom{n}{k}}=\frac{1}{2\pi i}\int_{C} \sqrt{f(z)}\frac{\pi}{\sin \pi z}dz,$$ where $C$ is any contour in  $-1<Re(z)<n+1$ which winds once about each integer $0,1,\ldots,n$ and never about any other integer.
Suppose we let $C=C_{M}$ be the rectangle formed by the lines $Re(z)=-\frac{1}{2}$, $Re(z)=n+1/2$, and $Im(z)=\pm M.$  Then
$$\int_{C_M} \sqrt{f(z)}\frac{\pi}{\sin \pi z}dz=\int_{C_{M}}\frac{\sqrt{\pi}\,dz}{\sqrt{z(1-z)(1-z/2)\cdots(1-z/n)\sin \pi z}}$$
and letting $M\rightarrow \infty$, we conclude that
$$\sum_{k=0}^{n}(-1)^{k} \sqrt{\binom{n}{k}}=\frac{1}{2\pi i}\left[\int_{-1/2+i \infty}^{-1/2- i \infty}+\int_{n+1/2-i \infty}^{n+1/2+i \infty}\sqrt{f(z)}\frac{\pi}{\sin \pi z}dz\right].$$
Since the integrand is invariant (aside from a $\pm$ sign) under the substotution $z\rightarrow n - z$, we need only estimate the first integral.  Now, when $Re(z)=-\frac{1}{2}$,
\begin{align*}
|z(1-z)(1-z/2)\cdots(1-z/n)|
&\ge \frac{1}{2}\left(1+\frac{1}{2}\right) \left(1+\frac{1}{4}\right)\cdots\left(1+\frac{1}{2n}\right) \\
&\ge\frac{1}{2}\sqrt{1+1}\sqrt{1+1/2}\cdots\sqrt{1+1/n}\\
&=\frac{\sqrt{n+1}}{2}.
\end{align*}
and so the first integral is bounded by
$$\frac{1}{\sqrt{2\pi}\sqrt[4]{n+1}}\int_{Re(z)=-1/2}|\frac{dz}{\sqrt{\sin\pi z}}|\le\frac{A}{\sqrt[4]{n}}.$$
Hence $$\sum_{k=0}^{n}(-1)^{k} \sqrt{\binom{n}{k}}\rightarrow0$$ as $n\rightarrow\infty.$
[The problem seems to have originated with D. J. Newman, who posed this as Problem 81-7 in the SIAM Review.  A more complicated proof, as well as a discussion of generalizations, appears on pp. 155-156 of Problems in Applied Mathematics: Selections from SIAM Review edited by Murray Klamkin (SIAM, 1990).  The proof given above can be found on pp. 151-152 of Complex Analysis (Second Edition) by Bak and Newman (Springer, 1997).]
A: The terms for $n$ odd are zero by cancelling the binomial at $k$ and $n-k$ since the signs are opposite. 
When $n$ is even, the terms do not seem to have a limit.
