A calculus problem by A Former Brilliant Member

Let's Concentrate to simplify $S_n$ first.

$S_n = \dfrac{2017}{2} + \displaystyle\sum_{r=1}^{2014} \dfrac{-r}{(2017-r)(2016-r)} \\ S_n = \dfrac{2017}{2} + \displaystyle\sum_{r=1}^{2014} \dfrac{(2017-r)-(2017)}{(2017-r)(2016-r)} \\ S_n = \dfrac{2017}{2} + \displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)}-2017\displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)(2017-r)} \\ S_n = \dfrac{2017}{2} + \displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)}-2017\displaystyle\sum_{r=1}^{2014} \dfrac{(2017-r)-(2016-r)}{(2016-r)(2017-r)} \\ S_n = \dfrac{2017}{2} + \displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)}-2017\left( \displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)}-\displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2017-r)} \right) \\ S_n = \cancel{\dfrac{2017}{2}} + \displaystyle\sum_{r=1}^{2014} \dfrac{1}{(2016-r)}-\cancel{\dfrac{2017}{2}}+\dfrac{2017}{2016} \\ S_n = \dfrac{2017}{2016} + \left( \dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\cdots+\dfrac{1}{2} \right) \\ S_n = \dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\cdots+\dfrac{1}{2}+\dfrac{1}{1}$

$\text{Now , } \dfrac{1-x^{n}}{1-x} = 1+x+x^2+x^3+\cdots+x^{n-1} \\ \displaystyle \int_{0}^{1} \dfrac{1-x^{n}}{1-x} dx = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{n}$

Comparing the Two results , we get $n=2016$

Answer : $\dfrac{2016}{2^{5} \cdot 3^{2}} = 7$

Relevant wiki: Telescoping Series - Sum

The Definition on the right is simply the sum of harmonic series upto n terms which obviously diverges and we cant find its exact form. (which can be seen by visualising that function as the sum of GP 1+x+x^2......+x^n-1)

The series on left inside the bracket can be written as sigma from 1 to 2014 of (2015-r)/(r+1)(r+2)

By partial fractions we can break it into two telescopic series and one harmonic series like

2015(1/r+1 - 1/r+2)-1/(r+2)+(1/r+1-1/r+2)

The first and third are brackets are readily evaluated using the standard telescopic series while the middle series is a harmonic series

On simplification of first and third series and combining it with the harmonic series the value of n comes out to be 2016