Adiós 2015

Nice problem!

$f(n)=\sum_{k=0}^{n}f(k)-\sum_{k=0}^{n-1}f(k)=2016\frac{n+1}{n+2}-2016\frac{n}{n+1}=\frac{2016}{(n+1)(n+2)}$ so $f(2014)=\frac{1}{2015}$ and the reciprocal we seek is $\boxed{2015}$

$\begin{aligned} \sum_{k=0}^{2014} f(k) & = f(0)+f(1)+ \ldots +f(2014)\\ & = 2016\frac{(2014+1)}{(2014+2)}\\ & = 2015 \end{aligned}$ $\begin{aligned} \sum_{k=0}^{2013} f(k) & = f(0)+f(1)+ \ldots +f(2013)\\ & = 2016\frac{(2013+1)}{(2013+2)}\\ & = \frac{2016 \cdot 2014}{2015} \end{aligned}$ $\begin{aligned} f(2014) & = \sum_{k=0}^{2014} f(k) - \sum_{k=0}^{2013} f(k)\\ & = 2015 - \frac{2016 \cdot 2014}{2015}\\ & = 2015 - \frac{(2015+1)(2015-1)}{2015}\\ & = \frac{2015^2-(2015^2-1^2)}{2015}\\ & = \frac{1}{2015} \end{aligned}$

$(f(2014))^{-1} = (\frac{1}{2015})^{-1} = 2015$