Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum\limits_n \frac{1}{n+f(n)} $ is convergent? Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum_1 ^\infty \frac{1}{n+f(n)} $ is convergent?
I could not solve this. I tried to proceed in following lines:
1) Tried to provide a contradiction:
First let $n \sim m$ iff $\exists k \in \mathbb{Z}$ such that $f^k(n)=m$. This is an equivalence relation. Consider the orbits. For the finite orbits we can compare the series to $\sum_1^\infty \frac{1}{n+n}$. But then I could not figure out how to proceed for infinite orbits.
2) Tried to prove that there is some function:
Let $\{k_n\}$ be a subsequence of $\mathbb{N}$ such that $\sum_0^\infty \frac{1}{k_n}$ converges. Set $f(n)=k_{n}, \forall n \in \mathbb{N}\setminus\{k_n\}$. Then images of each $n$ which are not in the subsequence $k_n$ is defined. Now we have to define images of each $k_n$.
Define $f(k_n)=n$ $ \forall n \in \mathbb{N}\setminus \{k_n\}.$.
Could not proceed further. I think My second attempt was going in right direction. My plan was use the fact that all elements here are positive and to construct the function $f$ in such a manner that $\forall n\in \mathbb{N}$ either $n$ or $f(n)$ is in $\{k_n\}$.
 A: 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 $f$.
A: 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. 
