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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?

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It is reasonable to believe that behaviour for odd, even $n$ is not too dissimilar, and for odd it is always $0$. –  André Nicolas Jun 21 '12 at 9:13
    
One way is to consider integral: $$\frac{1}{2i} \int_\gamma \sqrt{ \frac{\Gamma (n+1)}{\Gamma (n+1 - z) \Gamma (z+)}} \frac{1}{\sin \pi z} \, dz$$ where $\gamma$ is a rectangle with corners in $-\frac{1}{2} \pm ib$ and $n+\frac{1}{2} \pm ib$ than take limit $n, b \to +\infty$ which requires some manipulations along the road. –  qoqosz Jun 21 '12 at 10:28
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The pieces of a complete answer can be collected starting from here. –  Did Jun 23 '12 at 21:10

2 Answers 2

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 $Re(z)=-\frac{1}{2}$,

$|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} $

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).]

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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.

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