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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?

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    $\begingroup$ I remember doing such a question before. Just complete the square in numerator after lcm etc. It's the most motivating step to me. $\endgroup$ – Mann Jul 14 '15 at 12:18
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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

$$\sum_{k = 1}^n \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = \sum_{k = 1}^n \left(1 + \frac{1}{k} - \frac{1}{k+1}\right) = n + 1 - \frac{1}{n+1}.$$

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The expression under the square root is $$\frac {k^4+2k^3+3k^2+2k+1} {k^2 (k+1)^2} = \frac {(1+k+k^2)^2} {k^2 (k+1)^2}$$ So your sum becomes $$\sum \limits _{k=1} ^n \frac {1+k+k^2} {k (k+1)} = \sum \limits _{k=1} ^n \left(1 + \frac 1 {k (k+1)} \right) = n + \sum \limits _{k=1} ^n \left(\frac 1 k - \frac 1 {k+1}\right) = n + 1 - \frac 1 {n+1}$$.

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