Let $a_n$ and $b_n$ be two strictly increasing sequences of natural numbers,with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty$ and $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$

.When I was an undergraduate student, i had in mind the ''conjecture'' that using the above sequences it is impossible to have $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=c, c\in R$.

Surprisingly enough, i believe we can find a counterexample:
We can find two sequences $a_n,b_n$ with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty,$ $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$ but $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=1$ (or some other constant ,but this is a specific example)

Let $a_n+b_n=2^n$

The idea is to increase the sum of the first series by a number aproximately
$ln2$ between the $2^{2^{.^{.{^2}}}}$

($n-2$ exponents) and $2^{2^{.^{.{^2}}}}$ ($n$ exponents) term. The same idea will also run for the second series.

$a_1=b_1=1\\a_2=b_2=2\\a_3=5,b_3=3\\a_4=12,b_4=4$

because $a_4$ is ''too big'' we will increase $a_n$ only by $1$ for the next $12$ terms,and increase significally $b_n$ in the following way:

$a_5=13 , b_5=19,a_6=14, b_6=50 ...$ and so on ,until we reach $ a_{16}=24 ,b_{16}=65512 $$.$ Again we will switch the growth of $b_n$ and $a_n,$ adding only $1$ to $b_n$ for the next $65512$ terms but increase significally $a_n$ in order to sum $2^n$. So , we will reach at $2^{2^{2^{2}}}=2^{65536}=a_{65536}+b_{65536},$ with $a_{65536}=2^{65536}-131024$ and $b_{65536}=131024$ and then continue in the same way.

It easy to see that when we reach $2^{2^{.^{.{^2}}}}$ ($2n$ exponents) both $\sum{\frac{1}{a_n}}$ and $\sum{\frac{1}{b_n}}$ will be at least $nln2$ and so, conclude that both series reach infinity.

We could define a set $A$ of *pairs* of $(\sum{\frac{1}{a_n}}$,$\sum{\frac{1}{b_n}})$ both being divergent, but with the property that $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}$ is convergent. I believe that this will define the ''pair of slowest divergent series''

Note:we use $''ln2''$ because $\frac{1}{n+1}+...+\frac{1}{2n}$ has limit $ln2$