# An easy method?

Jan 21, 2018

Hint: $\frac{1}{\sqrt{n} + \sqrt{n + 1}} = \text{what}$?

#### Explanation:

Rationalize the denominators:

$\frac{1}{\left(\sqrt{n} + \sqrt{n + 1}\right)} \cdot \frac{\left(\sqrt{n} - \sqrt{n + 1}\right)}{\left(\sqrt{n} - \sqrt{n + 1}\right)} = \frac{\sqrt{n} - \sqrt{n + 1}}{n - \left(n + 1\right)}$

$= \sqrt{n + 1} - \sqrt{n}$

So the sum above is equivalent to

$\left(\sqrt{2} - 1\right) + \left(\sqrt{3} - \sqrt{2}\right) + \left(\sqrt{4} - \sqrt{3}\right) + \cdot \cdot \cdot + \left(\sqrt{8} - \sqrt{7}\right) + \left(\sqrt{9} - \sqrt{8}\right)$
Which is $- 1 + \sqrt{9} = 2$