<s>[INST] 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?
 [/INST] 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.
</s>