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? [duplicate]

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.

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.