Provide example of two series that both diverge but $\sum\min\{a_n,b_n\}$ converges I've posted the solution for this problem and I'm trying to understand this. 
In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $\frac{1}{n^2}$ and $b_n$ to be $\frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $\frac{1}{n^2}$ and $a_n$ to be $\frac{1}{810000 ^2}$ for the next $810000 ^2$ terms?   (because $900^2$ is $810000$)
And why do we have to add one to the sum of partial sums?

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $\sum a_n,\sum b_n$ both diverge, but $\sum \min\{a_n,b_n\}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $\min\{a_n,b_n\} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $\sum a_n,\sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.
To find how many terms we add again, think about the pattern
\begin{align*}
(1+1) - 1 &= 1^2 \\
(5 + 1) - 2 &= 2^2 \\
(30 + 1) - 6 &= 5^2 \\
(930 + 1) - 31 &= 30^2 \\
(865830 + 1) - 931 &= 930^2 \\
(x + 1) - 865831 &= 865830^2 \\
\dotsb
\end{align*}
