Provide example of two series that both diverge but $\sum\min\{a_n,b_n\}$ converges

Question

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?

Answer 1

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*}

Answer 2

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 $\frac{1}{n^2}$.