For $a_n,b_n\uparrow$ and $\sum \frac{1}{a_n}$, $\sum \frac{1}{b_n}$ divergent is the series $\sum \frac{1}{a_n+b_n}$ also divergent? Let $a_n$ and $b_n$ are strictly increasing to $+\infty$ sequences such that the series $\sum \frac{1}{a_n}$ and $\sum \frac{1}{b_n}$ are divergent. Is it true that the series $\sum \frac{1}{a_n+b_n}$ is also divergent? 
At first sight it looks true, so I tried to prove that, but after some failed attempts I start to believe now that there might exist a counterexample. Any thoughts?
 A: For $r \geqslant 0$, let $k_r = 2^{2^r}$. Let
$$\begin{align}
a_n &= k_{2r+2} - \frac{1}{n},\text{ for } k_{2r} \leqslant n < k_{2r+2};\\
b_n &= k_{2r+1} - \frac{1}{n},\text{ for } k_{2r-1} \leqslant n < k_{2r+1};
\end{align}$$
for $n \geqslant 4$, and choose $a_n, b_n$ fairly arbitrarily for $n < 4$.
Then
$$\sum_{n=k_{2r}}^{k_{2r+2}-1} \frac{1}{a_n} > \frac{k_{2r+2}-k_{2r}}{k_{2r+2}} > \frac{1}{2},$$
so $\sum \frac{1}{a_n}$ diverges. Analogously, $\sum \frac{1}{b_n}$ diverges.
But, we have $a_n > b_n$ for $k_{2r} \leqslant n < k_{2r+1}$, and $b_n > a_n$ for $k_{2r+1} \leqslant n < k_{2r+2}$, so
$$\sum_{n=k_{2r}}^{k_{2r+2}-1} \frac{1}{a_n + b_n} < \frac{k_{2r+1}-k_{2r}}{k_{2r+2}} + \frac{k_{2r+2}-k_{2r+1}}{k_{2r+3}} < \frac{k_{2r+1}}{k_{2r+2}} + \frac{k_{2r+2}}{k_{2r+3}},$$
and
$$\frac{k_r}{k_{r+1}} = 2^{2^r-2^{r+1}} = 2^{-2^r} = \frac{1}{k_r},$$
so
$$\sum_{r=1}^\infty \frac{1}{k_r} < \infty$$
and
$$\sum_{n=1}^\infty \frac{1}{a_n+b_n}$$
converges.
A: This is not true. Note that ${1 \over a_n + b_n} \leq {1 \over \max({a_n,b_n})} = \min({{1 \over a_n},{1 \over b_n}})$. So it suffices to find an example where $\sum_n \min({{1 \over a_n},{1 \over b_n}})$ converges. 
To do this, you can choose $N_1 < N_2 < ...$ such that 
for $n$ in $[N_i, N_{i+1}]$ one of ${1 \over a_n}$ and ${1 \over b_n}$ is equal to $2^{-n}$ and the other is almost constant on $[N_i, N_{i+1}]$. If $N_{i+1}$ is large enough relative to $N_i$, then the sum of the terms of the almost-constant sequence can be made greater than $1$ if they decrease slowly enough. 
Then on $[N_{i+1}, N_{i+2}]$ you switch roles; the other of ${1 \over a_n}$ and ${1 \over b_n}$ is equal to $2^{-n}$ and the one that was formerly  equal to $2^{-n}$ is now nearly constant making sure the sequence is strictly decreasing. You then switch roles again on $[N_{i+2}, N_{i+3}]$, and so on ad infinitum. 
Since $\sum_n \min({{1 \over a_n},{1 \over b_n}}) \leq \sum_n 2^{-n} = 1$, the sum $\sum_n {1 \over a_n + b_n} $ is finite. But because each of the original sums is greater than $1$ on every other block, the both diverge.
A: $\mathbf{Edit:}$ This is wrong. I missed the strictly increasing part, leaving it in case it helps for an answer.
No. Hint: Suppose $a_n$ is like $n$ for even $n$, but is like $n^2$ for odd $n$. Now, suppose the opposite for $b_n$ (it is like $n^2$ for even $n$ but like $n$ for odd $n$).
A: This is false, although I dont have a concrete counterexample I have a construction, of a general counterexample.
Define $a_n$ and $b_n$ in blocks. To make $\frac{1}{a_n}$ diverge, if we have already defined $a_{n_0}$ take the sequence $$\frac{1}{a_{n_0}+1}+ \frac{1}{a_{n_0}+2}+ 
\frac{1}{a_{n_0}+3}+ 
\frac{1}{a_{n_0}+4}+\cdots$$  and continue this for as long a you like building up a large number, now chose $b_n$ corespondingly so that $\frac{1}{a_n+b_n}$ is very small, bounded by $\frac{1}{2^n}$ say.
After you have built up a large number on the $a$ side switch the roles and build up a large value on the $b$ side and chose $a$'s to make the sum converge. Now just keep alternating this.
A: There exists a counterexample.
We construct $a_n,b_n$ inductively in blocks of increasing size. Each block will contribute $\simeq 1$ to $\sum_n 1/a_n,\ \sum_n 1/b_n$ but the contribution to $\sum_n \frac{1}{a_n+b_n}$ will be negligible.
Suppose that $a_{n_0}, b_{n_0}$ are the first terms not yet contructed, and let $N \geq a_{n_0-1},b_{n_0-1}$ be an integer. We set $a_{n} = N$ for $n_0 \leq n < n_0 + N$ and for larger $n \leq n_0 + N^2$ we assign some insanely large value. We set $b_{n} = N^2$ for $n_0 \leq n < n_0 + N^2$. The newly constructed block ends at $n_0 + N^2$.
The entries we have just contructed contribute about $1$ to each of $\sum_n 1/a_n,\ \sum_n 1/b_n$. On the other hand, they contribute about $1/N$ to $\sum_n \frac{1}{a_n+b_n}$. If we set $N$ to be sufficiently large (e.g. if we are looking at $i$-th block, we can require that $N > 2^i$) we find that  $\sum_n \frac{1}{a_n+b_n}$ converges.
Edit: If $a_n,b_n$ are supposed to be strictly increasing, just add something like $1-1/n$ to the sequences constructed above.
