Because, it's the year 2015!

$let$ $t=n^{2015}$

The given expression becomes

$\large{\pi \div \displaystyle \lim_{t\to \infty} \left [t^2 \left (\frac{\arctan(t)}{t}-\frac{\arctan (t+1)}{t+1} \right) \right]}$

Simplifying and using

$\arctan(a)-\arctan(b)=\arctan \left(\frac{a-b}{1+ab} \right )$

$\large{ \displaystyle \lim_{t\to \infty} \left [\frac{t^2}{t(t+1)} \left (t(\arctan \left(\frac{t-(t+1)}{1+t(t+1)} \right))+\arctan t \right) \right]}$

simplifying

$\large{\displaystyle \lim_{t\to \infty} \left[\frac{1}{1+\frac{1}{t}}\left(t(\arctan \left(\frac{-1}{1+t(t+1)} \right))+\arctan t \right) \right]}$

Applying the limit we get

$\large{1(t\times \arctan (0))+\arctan t}$ where $t\to \infty$

$\large{\boxed{\frac{\pi}{2}}}$