Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$

Question

Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$

Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$

A lot of these terms cancel, and I think that the first and last terms will be left.

I feel like I'm close, but missing something.

Answer 1

We can rewrite the sum as:

$$\sum_{n=1}^{99} \frac{2}{\sqrt{n} + \sqrt{n+2}}\frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}= \sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}$$

It's a telescopic sum, where most of the terms are cancelled.

We can expand it:

$$\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}= (\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})+\dots+(\sqrt{101}-\sqrt{99})=\sqrt{100}+\sqrt{101}-\sqrt{1}-\sqrt{2}$$

Answer 2

HINT:

Note that we have

$$\frac1{\sqrt n+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt n}{2}$$

Now telescope.

Answer 3

Let $ m \in\mathbb{N}^{*} : $

\begin{aligned}\sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n}\right)}\\ &=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n+1}\right)}+\sum_{n=1}^{m}{\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sqrt{m+2}-\sqrt{2}+\sqrt{m+1}-1\end{aligned}