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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 $0$.2012-06-21
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    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.2012-06-21
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    The pieces of a complete answer can be collected starting from [here](http://mathoverflow.net/questions/85013/alternating-sum-of-square-roots-of-binomial-coefficients).2012-06-23

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