Telescoping Series

$\begin{aligned} S & = \frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \frac 1{\sqrt 3 + \sqrt 4} + ... + \frac 1{\sqrt{2016} + \sqrt {2017}} \\ & = \frac {\sqrt 2 - \sqrt 1}{(\sqrt 2 + \sqrt 1)(\sqrt 2 - \sqrt 1)} + \frac {\sqrt 3 - \sqrt 2}{(\sqrt 3 + \sqrt 2)(\sqrt 3 - \sqrt 2)} + \frac {\sqrt 4 - \sqrt 3}{(\sqrt 4 + \sqrt 3)(\sqrt 4 - \sqrt 3)} + ... + \frac {\sqrt {2017} - \sqrt {2016}}{(\sqrt {2017} + \sqrt {2016})(\sqrt {2017} - \sqrt {2016})} \\ & = \frac {\sqrt 2 - \sqrt 1}{2-1} + \frac {\sqrt 3 - \sqrt 2}{3-2} + \frac {\sqrt 4 - \sqrt 3}{4-3} + ... + \frac {\sqrt {2017} - \sqrt {2016}}{2017-2016} \\ & = \frac {\sqrt 2 - \sqrt 1}1 + \frac {\sqrt 3 - \sqrt 2}1 + \frac {\sqrt 4 - \sqrt 3}1 + ... + \frac {\sqrt {2017} - \sqrt {2016}}1 \\ & = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + ... + \sqrt {2017} - \sqrt {2016} \\ & = \cancel{\sqrt 2} - \sqrt 1 + \cancel{\sqrt 3} - \cancel{\sqrt 2} + \cancel{\sqrt 4} - \cancel{\sqrt 3} + ... + \sqrt {2017} - \cancel{\sqrt {2016}} \\ & = \boxed{\sqrt{2017} - 1} \end{aligned}$

I'm just going to make this simple, because the LaTeX for this would take quite a long time.

First, rationalize the denominators. We can see here that there is one negative $\sqrt{2}$ value and one positive $\sqrt{2}$ value. This goes for all radicals up to $\sqrt{2016}$ . Therefore, these values cancel out. After this simplification is performed, $\sqrt{2017}$ only appears once (as a positive value) and in the beginning, $\sqrt{1}$ appears as a negative value (when the entire expression is simplified). Simplifying, we subtract $\sqrt{2017}-\sqrt{1}=\sqrt{2017}-1$

For any $k\geq 0$ we may write $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1} - \sqrt{k}}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k+1} - \sqrt{k}\right)} = \frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k} = \sqrt{k+1}-\sqrt{k}$

Applying this to each term of the given series allows us to rewrite it as $\left(\sqrt{2} - \sqrt{1}\right) + \left(\sqrt{3} - \sqrt{2}\right) + \left(\sqrt{4} - \sqrt{3}\right) + \cdots + \left(\sqrt{2016} - \sqrt{2015}\right) + \left(\sqrt{2017} - \sqrt{2016}\right) = \sqrt{2017} - \sqrt{1} = \boxed{\sqrt{2017}-1}$