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