How to find the sum $\sum_{k = 1}^n \frac{1}{(k+1)\sqrt{k}+k\sqrt{(k+1)}}$?

Question

How do you find the sum $$\sum_{k = 1}^n \frac{1}{(k+1)\sqrt{k}+k\sqrt{(k+1)}}?$$ Sorry for not provide any of my idea since I have no idea about what to do.

I hope everyone here could help me. Thank you

Answer 1

As suggested in the comments, we have that

$$ \frac{1}{(k+1)\sqrt{k}+k\sqrt{(k+1)}}\cdot \frac{(k+1)\sqrt{k}-k\sqrt{(k+1)}}{(k+1)\sqrt{k}-k\sqrt{(k+1)}}= \frac{(k+1)\sqrt{k}-k\sqrt{(k+1)}}{k^2+k}=$$

$$=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$

and we obtain a telescoping sum.