TITLE: Summation stuck under radical sign
QUESTION [5 upvotes]: I am trying to evaluate the following sum, but I'm unable to solve it in any general way.
$$S=\sum_{k=1}^n\sqrt{1+\frac{1}{(k)^2}+\frac{1}{(k+1)^2} }$$
How can I do it?

REPLY [17 votes]: A different method:
\begin{align}1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} &= 1 + \frac{2}{k(k+1)} + \left[\frac{1}{k^2} - \frac{2}{k(k+1)} + \frac{1}{(k+1)^2}\right]\\
&= 1 + 2\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{k} - \frac{1}{k+1}\right)^2\\
&= \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)^2,
\end{align}
and thus 
$$\sum_{k = 1}^n \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = \sum_{k = 1}^n \left(1 + \frac{1}{k} - \frac{1}{k+1}\right) = n + 1 - \frac{1}{n+1}.$$