Question

Sol. [Key Concept: Rationalise $\frac{1}{(k + 1)\sqrt{\ }k + k\sqrt{k + 1}} = \frac{1}{\sqrt{\ }k} - \frac{1}{\sqrt{k + 1}}$ — telescopes.]

$$\sum_{k = 1}^{n}\mspace{2mu}\frac{1}{\sqrt{k}\sqrt{k + 1}(\sqrt{k + 1} + \sqrt{k})} = \sum_{k = 1}^{n}\mspace{2mu}\frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k}\sqrt{k + 1}}$$

$$= \sum_{k = 1}^{n}\mspace{2mu}\left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}} \right) = 1 - \frac{1}{\sqrt{n + 1}}$$