First, expand $\displaystyle\sum_{k=2}^{\infty}f(k)$: $$ \begin{array}{ccccccccccc} f(2) &=& \frac{1}{2^2} &+& \frac{1}{3^2} &+& \frac{1}{4^2} &+& \cdots &+& \frac{1}{2016^2}\\ f(3) &=& \frac{1}{2^3} &+& \frac{1}{3^3} &+& \frac{1}{4^3} &+& \cdots &+& \frac{1}{2016^3}\\ f(4) &=& \frac{1}{2^4} &+& \frac{1}{3^4} &+& \frac{1}{4^4} &+& \cdots &+& \frac{1}{2016^4}\\ &\cdots \end{array} $$ Next, add these terms by column. Each forms an infinite geometric sequence. $$ \begin{array}{ccccc} \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \cdots &=& \frac{1}{2^2}\times\frac{1}{1-\frac{1}{2}}&=&\frac{1}{1\times 2}\\ \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots &=& \frac{1}{3^2}\times\frac{1}{1-\frac{1}{3}}&=&\frac{1}{2\times 3}\\ \frac{1}{4^2} + \frac{1}{4^3} + \frac{1}{4^4} + \cdots &=& \frac{1}{4^2}\times\frac{1}{1-\frac{1}{4}}&=&\frac{1}{3\times 4}\\ \cdots\\ \frac{1}{2016^2} + \frac{1}{2016^3} + \frac{1}{2016^4} + \cdots &=& \frac{1}{2016^2}\times\frac{1}{1-\frac{1}{2016}}&=&\frac{1}{2015\times 2016}\\ \end{array} $$ Therefore, the sum equals \begin{align*} &\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots+\frac{1}{2015\times 2016}\\ =& \Big(\frac{1}{1}-\frac{1}{2}\Big)+\Big(\frac{1}{2}-\frac{1}{3}\Big)+\cdots +\Big(\frac{1}{2015}-\frac{1}{2016}\Big)\\ =& \boxed{\frac{2015}{2016}} \end{align*}