Simplify It

Note that for any positive integer $n$ we have that

$1 + \dfrac{1}{n^{2}} + \dfrac{1}{(n + 1)^{2}} = \dfrac{n^{2}(n + 1)^{2} + (n + 1)^{2} + n^{2}}{n^{2}(n + 1)^{2}} = \dfrac{n^{4} + 2n^{3} + 3n^{2} + 2n + 1}{n^{2}(n + 1)^{2}} = \dfrac{(n^{2} + n + 1)^{2}}{n^{2}(n + 1)^{2}}$ ,

the square root of which is $\dfrac{n^{2} + n + 1}{n(n + 1)} = 1 + \dfrac{1}{n(n + 1)} = 1 + \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$ .

So the given sum, namely $\displaystyle\sum_{n=1}^{2015}\left(1 + \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)\right)$ telescopes to

$2015 + \left(1 - \dfrac{1}{2016}\right) = \boxed{2016 - \dfrac{1}{2016}}$ .