Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
The pieces of a complete answer can be collected starting from here. – Did Jun 23 '12 at 21:10

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).]

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.