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

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?

You can turn the sum into two Riemann sums each of which converges to $\frac \pi 2$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}} = \sum_{k=1}^{n}\frac{n+1-k + k}{\sqrt{n+1-k}\cdot\sqrt{k}}\cdot \frac 1{n+1}$$
$$=\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{n+1-k}{k}} + \frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{k}{n+1-k}}$$
$$= \underbrace{\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{1}{\frac{k}{n+1}}-1}}_{\stackrel{n\to \infty}{\rightarrow}\int_0^1\sqrt{\frac 1x-1}dx=\frac \pi2} +  \underbrace{\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{1}{\frac{1}{\frac{k}{n+1}}-1}}}_{\stackrel{n\to \infty}{\rightarrow}\int_0^1\sqrt{\frac{1}{\frac 1x -1}}dx=\frac \pi2}$$
