TITLE: Divergence of $\sum\limits_n1/\max(a_n,b_n)$
QUESTION [11 upvotes]: Given two positive and (strictly) monotone increasing sequences, $a_n$ and $b_n$ such that $\displaystyle \sum_{n=1}^{\infty}\dfrac1{a_n}$ and $\displaystyle \sum_{n=1}^{\infty}\dfrac1{b_n}$ diverge, does $\displaystyle \sum_{n=1}^{\infty}\dfrac1{\max(a_n,b_n)}$ diverge?

For sequences, where $a_n > b_n$ or $a_n < b_n$ eventually, the result follows immediately. But I am unable to see what happens when $a_n$ and $b_n$ take turns to overtake each other as $n \to \infty$?
Essentially this question stems from the question, Osgood condition , where I wanted to consider the function $F(x) = g(x) + h(x)$ and lower bound $\displaystyle \int\dfrac{dy}{F(y)}$ by $\displaystyle \int \dfrac{dy}{2\max(h(y),g(y))}$ in an attempt to show that $\displaystyle \int\dfrac{dy}{F(y)}$ diverges.

REPLY [7 votes]: This question is asked now and then. The answer is that the series with the maxima in the denominators can converge despite both initial series diverge and the reason is exactly as you put it: $a_k$ and $b_k$ can overtake each other infinitely many times and accumulate large sums when at rest from the work as maxima. Formalizing it isn't hard and did did an excellent job in this respect. The question is whether we can make it "obvious", i.e., to see it in our heads without touching pen or paper at all. Of course, we can talk about decreasing sequences $a_k,b_k>0$ and consider $\sum_k\min(a_k,b_k)$. Now just imagine two people walking. The condition is that neither of them is allowed to increase his speed at any time and that the slowest one carries a stick, which is magically teleported to the other person when he becomes slower. The question is whether both people can walk to infinity while the stick will travel only finite distance on foot. Now the strategy should become clear. Fix any upper bound $v$ for speeds and any upper bound $d$ for the distance the stick is allowed to travel on foot. Let the first person walk at speed $v$ until he goes the distance $1$. The second person should crawl slowly (but steadily) all that time at the speed $dv/2$, so he travels only distance $d/2$ with the stick. At that moment, the first person slows down enormously to the speed $d^2v/4$, so while the second person continues to go at the speed $dv/2$ and moves distance $1$, the first person (who now has the stick) goes only $d/2$. By the end of this cycle, both people moved by at least $1$ but the stick traveled only distance $d$ on foot. We also end up with the new bound for the speed, which is $vd^2/4$. Now repeat the cycles making the allowed distances for the stick smaller and smaller so that the corresponding series converges. We do not care how fast the speed bounds decay because no matter what the bound is, we still can cover the unit distance in some finite time. During each cycle, each person moves by distance $1$ or more, so both ultimately walk away. However, the stick goes only finite distance on foot.