Your sum can be rewritten as
$ \sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \sum_{f=2}^{\infty} \frac{1}{f^s} - \sum_{g=2}^{\infty} \frac{2}{(g+1)g-2}. $
Now, the right-hand sum can be evaluated explicitly since it telescopes. Indeed,
$ \begin{align*} \sum_{g=2}^{\infty} \frac{2}{(g+1)g-2} &= 2\sum_{g=2}^{\infty} \frac{1}{(g+1)g-2} \\ &= 2\sum_{g=2}^{\infty} \frac{1}{(g-1)(g+2)} \\ &= \frac{2}{3} \sum_{g=2}^{\infty} \left(\frac{1}{g-1} - \frac{1}{g+2}\right) \\ &= \frac{2}{3} \left(1 \color{red}{- \frac{1}{4}} + \frac{1}{2} \color{blue}{- \frac{1}{5}} + \frac{1}{3} \color{violet}{- \frac{1}{6}} \color{red}{+ \frac{1}{4}} \color{violet}{- \frac{1}{7}} \color{blue}{+ \frac{1}{5}} \color{violet}{- \frac{1}{8}} + \cdots\right) \\ &= \frac{2}{3}\left(1 + \frac{1}{2} + \frac{1}{3}\right) \\ &= \frac{11}{9}, \end{align*} $
where the terms in $\color{red}{\text{red}}$ cancel with each other, the terms in $\color{blue}{\text{blue}}$ cancel with each other, and the terms in $\color{violet}{\text{violet}}$ cancel with something that isn't shown.
Then, since
$ \sum_{f=2}^{\infty} \frac{1}{f^s} = \sum_{f=1}^{\infty} \frac{1}{f^s} - 1 = \zeta(s) - 1, $
where $\zeta$ is the Riemann zeta function, we see that your sum is equal to
$ \sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \zeta(s) - 1 - \frac{11}{9} = \zeta(s) - \frac{20}{9}. $
The Riemann zeta function converges for all $s$ with $\Re(s) > 1$, so the same is true for your sum.