Saturday, 15 October 2016

limits - Evaluate $limlimits_{n to infty}sum_{k=1}^{n}frac{1}{sqrt{n+1-k}cdotsqrt{k}}$


$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}$$




Maybe we can believe with assurance that



$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}\approx\sum_{k=1}^{n}\frac{1}{\sqrt{n-k}\cdot\sqrt{k}}=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{\frac{k}{n}-\left(\frac{k}{n}\right)^2}}.$$
Thus, we can obtain
$$\lim_{n \to

\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}=\int_0^1 \frac{{\rm d}x}{\sqrt{x-x^2}}=\pi$$

This is true? If so, how to prove it rigorously?

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