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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?

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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*}

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Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $\frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series \begin{matrix} a_n = \bigl(1 & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \frac{1}{5^2} & \frac{1}{6^2} & \frac{1}{7^2} & \frac{1}{8^2} & \frac{1}{9^2} & \frac{1}{10^2} & \frac{1}{11^2} & \frac{1}{12^2} & \frac{1}{13^2} & \frac{1}{14^2} & \frac{1}{15^2} & \frac{1}{16^2} & \cdots &\bigr) \\ b_n = \bigl(1 & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \frac{1}{5^2} & \frac{1}{6^2} & \frac{1}{7^2} & \frac{1}{8^2} & \frac{1}{9^2} & \frac{1}{10^2} & \frac{1}{11^2} & \frac{1}{12^2} & \frac{1}{13^2} & \frac{1}{14^2} & \frac{1}{15^2} & \frac{1}{16^2} & \cdots &\bigr) \end{matrix} and modify them like so: \begin{matrix} a_n = \bigl(1 & \color{red}{\frac{1}{4}} & \color{red}{\frac{1}{4}} & \color{red}{\frac{1}{4}} & \color{red}{\frac{1}{4}} & \frac{1}{6^2} & \frac{1}{7^2} & \cdots & \frac{1}{29^2} & \color{red}{\frac{1}{900}} & \color{red}{\frac{1}{900}} & \color{red}{\frac{1}{900}} & \cdots & \color{red}{\frac{1}{900}} & \color{red}{\frac{1}{900}} & \frac{1}{930^2} & \cdots &\bigr) \\ b_n = \bigl(1 & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \color{red}{\frac{1}{25}} & \color{red}{\frac{1}{25}} & \color{red}{\frac{1}{25}} & \cdots & \color{red}{\frac{1}{25}} & \color{red}{\frac{1}{25}} & \frac{1}{31^2} & \frac{1}{32^2} & \cdots & \frac{1}{928^2} & \color{red}{\frac{1}{929^2}} & \color{red}{\frac{1}{929^2}} & \cdots &\bigr). \end{matrix} The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $\frac{1}{n^2}$.

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