Construct series Are there two non-negative, monotone sequences ${\{a_n\}}$ and ${\{b_n\}}$, s.t. $\sum{a_n}$ and $\sum{b_n}$ diverge, but $\sum{min{(a_n,b_n)}}$ converges?
I guess that the convergence speed must be carefully controlled, but I couldn't find such sequences. Thanks in advance for any help.
 A: Yes there are two such sequences. It is not easy to write an explicit formula for the general terms, though. Here is one of the possible constructions:


*

*Add $(2n)!$ terms to both series. The terms of $\{a_i\}$ should all equal $1\over(2n+1)!$, and those of $\{b_i\}$ should equal $1\over(2n+2)!$.

*Add $(2n+1)!$ terms to both series. The terms of $\{a_i\}$ should all equal $1\over(2n+3)!$, and those of $\{b_i\}$ should again equal $1\over(2n+2)!$.

*Increment $n$ and repeat.

A: This is possible. Let $d_1=1, d_{i+1}=d_i^2+d_i, i \ge 1$. Obviously, $d_i$ is strictly monotonically increasing. For any $n > 0$, let $f(n)$ be the index $i$ with $d_i \le n < d_{i+1}$.
Let's now define our series':
$$
a_n:=
\begin{cases} 
 \frac1{n^2} & \text{if } f(n) \text{ is even}\\ 
 \frac1{d_{f(n)}^2} & \text{if } f(n) \text{ is odd}\\ 
\end{cases}
$$
$$
b_n:=
\begin{cases} 
 \frac1{n^2} & \text{if } f(n) \text{ is odd}\\ 
 \frac1{d_{f(n)}^2} & \text{if } f(n) \text{ is even}\\ 
\end{cases}
$$
We subdivided the integers into intervals $[d_i,d_{i+1})$ and, with $n \in [d_i,d_{i+1})$ and $s(x)=\frac1{x^2}$, have the series take the value of $s(n)$ or $s(d_i)$, depending on $i$ being odd or even.
This shows that both $a_n,b_n$ are monotonically decreasing. We also have $\min(a_n,b_n)=\frac1{n^2}$, which means that $\sum \min(a_n,b_n)$ converges.
OTOH, for each interval $[d_i,d_{i+1})$ with odd $i$, we have
$$ \sum_{n=d_i}^{d_{i+1}-1} a_n = \sum_{n=d_i}^{d_{i+1}-1} \frac1{d_i^2} = (d_{i+1} - d_i)\frac1{d_i^2}= d_i^2\frac1{d_i^2} = 1,$$
because of the definition of $d_i$. So the series $\sum a_n$ has infinitely many disjoint finite parts that sum to 1, which means the sum diverges. The same argument (for even $i$) can be made for $\sum b_n$.
