Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum\limits_n \frac{1}{n+f(n)} $ is convergent?

Originally, I gave this example:

$$f (n)=\begin{cases}k,&\ \text { if } n=3^k , \text { with $k $ not a power of $2$}\\ 2^{n-1} , &\ \text { otherwise }\end{cases}$$ (the idea is to push the small numbers further and further down the road so that when they appear they are compensated by the $n $). Then $$ \sum_n\frac1 {n+f (n)}<\sum_{k}\frac 1 {3^k+k}+\sum_n\frac1 {n+2^{n-1}}<\infty. $$

And it is the right idea, but the problem is that such $f$ is not onto. For instance, $2^{26}$ is not in the range of $f$, because when $n=27$, we are using the other branch of $f$ to get $3$.

So we need to tweak the example slightly. Let $$ T=\{3^k:\ k\ \text{ is not a power of } 2\}=\{3^3,3^5,3^6,3^7,3^9,\ldots\} $$ and $$ S=\mathbb N\setminus T=\{1,\ldots,25,26,28,29,\ldots\}. $$ Write them as an ordered sequence, $T=\{t_1,t_2,\ldots\}$ and $S=\{s_1,s_2,\ldots\}$. Now define $$ f(n)=\begin{cases} \log_3 n,&\ \text{ if }\ n=t_k\\ \ \\ 2^{k-1},&\ \text{ if }\ n=s_k \end{cases} $$ One can check explicitly that $$ g(m)=\begin{cases} 3^m,&\ \text{ if $m$ is not a power of $2$}\ \\ \ \\ s_{k+1},&\ \text{ if }\ m=2^k \end{cases} $$ is an inverse for $f$.

Not a new solution, just writing to clarify for myself how the solution of @Martin Argerami: works.

We want $\sum_{n\in \mathbb{N}} \frac{1}{n + f(n)} < \infty$. Consider a covering $\mathbb{N} = A\cup B$. It's enough to have $\sum_{n\in A} \frac{1}{n + f(n)} $, $\sum_{n \in B} \frac{1}{n + f(n)}< \infty$. So it's enoough to find $A$, $B$ so that $$\sum_{n \in A} \frac{1}{f(n)}= \sum_{n \in f(A)} \frac{1}{n} < \infty \\ \sum_{n \in B} \frac{1}{n} < \infty$$

So it's enough to have $B$ so that $\sum_{n\in B} \frac{1}{n} < \infty$ and $f$ mapping $A = \mathbb{N} \backslash B$ to $B$. There are many possibilities here.