Wlog $f:\mathbb{N} \rightarrow \mathbb{N}$.
We consider the finite sum $\sum_{k=1}^N \frac{a_k}{f(k)}$. If necessary, we may rearrange $\{f(1),f(2),...,f(N)\}$ in non-decreasing order and apply the rearrangement inequality $$\sum_{k=1}^{N} \frac{a_k}{f(k)} \leq \sum_{k=1}^{N} \frac{a_k}{f(\sigma(k))}$$ where $\sigma$ is a permutation of the $N$ points $\{1,...,N\}$ s.t. $f(\sigma(n))$ is non-decreasing on $[1,N]$. Thus for every $N$ we may assume $f$ to be non-decreasing.
We can then estimate as in the following sequence of steps, since $a_k$ is decreasing \begin{align} \sum_{k=f(n)}^{f(N+1)-1}\frac{a_k}{f(k)} &= \sum_{\substack{k=f(n) \\ f(k)\leq k}}^{f(N+1)-1}\frac{a_k}{f(k)} + \sum_{\substack{k=f(n) \\ f(k)> k}}^{f(N+1)-1}\frac{a_k}{f(k)} = S_{f\leq k} + S_{f>k} \\ S_{f\leq k} &\leq \sum_{\substack{k=f(n) \\ f(k)\leq k}}^{f(N+1)-1}\frac{a_{f(k)}}{f(k)} \leq \sum_{\substack{k=1}}^{\infty}a_{f(k)} < \infty \\ S_{f>k} &= \sum_{\substack{k=n}}^{N} \sum_{\substack{m=f(k) \\ f(m)>m}}^{f(k+1)-1} \frac{a_{m}}{f(m)} \leq \sum_{\substack{k=n \\ \text{smallest } m \text{ s.t.} \\ f(m)>m \geq f(k)}}^{N} a_{f(k)} \, \frac{f(k+1)-f(k)}{f(m)} \\ &\leq \sum_{k=n}^N a_{f(k)} \leq \sum_{k=1}^\infty a_{f(k)} < \infty \, ,\end{align} since $f$ is non-decreasing and $f(m)>f(k)$, either $f(m)=f(k+1)$ or $f(m)>f(k+1)$. We can then take the limit $N\rightarrow \infty$.