# Divergence of a sequence

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.

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Two related questions: Slowing down divergence 2 and a follow-up. – commenter Oct 31 '12 at 8:24

The divergence of the series $\sum\limits_n1/a_n$ and $\sum\limits_n1/b_n$ does not imply the divergence of the series $\sum\limits_n1/\max\{a_n,b_n\}$.

Consider an increasing sequence $(N_k)$ of positive integers and define the sequences $(a_n)$ and $(b_n)$ on each interval $(N_k,N_{k+1}]$ as follows:

• If $N_{2k-1}\lt n\leqslant N_{2k}$ for some $k$, then $a_n=N_{2k-1}^2+n-N_{2k-1}$ and $b_n=n^2$.
• If $N_{2k}\lt n\leqslant N_{2k+1}$ for some $k$, then $a_n=n^2$ and $b_n=N_{2k}^2+n-N_{2k}$.

The sequences $(a_n)$ and $(b_n)$ are integer valued and increasing, and $\max\{a_n,b_n\}=n^2$ for every $n$ hence the series $\sum\limits_n1/\max\{a_n,b_n\}$ converges.

We now look for sufficient conditions on the sequence $(N_k)$ for the series $\sum\limits_n1/a_n$ and $\sum\limits_n1/b_n$ to diverge. Consider $$A_k=\sum\limits_{n=N_{k}+1}^{N_{k+1}}\frac1{a_n},\qquad B_k=\sum\limits_{n=N_{k}+1}^{N_{k+1}}\frac1{b_n},\qquad C_k=\frac{N_{k+1}-N_{k}}{N_{k}^2}.$$ Then $$A_{2k-1}=\sum\limits_{n=1}^{N_{2k}-N_{2k-1}}\frac1{N_{2k-1}^2+n},\qquad B_{2k}=\sum\limits_{n=1}^{N_{2k+1}-N_{2k}}\frac1{N_{2k}^2+n}.$$ The usual comparison with the integral of the function $x\mapsto1/x$ shows that $$A_{2k-1}\approx\log(1+C_{2k-1}),\qquad B_{2k}\approx\log(1+C_{2k}).$$ As soon as $\liminf C_k\gt0$, the series $\sum\limits_n1/a_n$ and $\sum\limits_n1/b_n$ diverge. Choosing $N_k=2^{2^k}$ for every $k$ yields $C_k\to1$, hence the proof is complete.

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Thanks. This is clear and neat. – user17762 Jan 13 '13 at 21:23

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.

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+1. Your stick going on foot is almost poetry... – Did Oct 31 '12 at 7:37

Here's my brute force example... based on convergence of 1/n versus divergence of 1/n^2.

The fact that $\sum \frac{1}{n}$ diverges, while $\sum \frac{1}{n^2}$ converges, can be used to construct strictly increasing integer sequences $a_n$ and $b_n$ for which each of $\sum \frac{1}{a_n}$ and $\sum \frac{1}{b_n}$ diverges, but nevertheless $\sum \frac{1}{max(a_n,b_n)}$ converges. To describe these sequences, let the number $k=1.5$ be called the "divergence bound". What we do is alternately put sequential numbers in one of the sequences, while putting squares in the other, making sure to put enough sequential numbers in each block in each case to exceed the divergence bound. Then a sequence of adjacent squares is placed in the other sequence, after which we switch the roles of the two sequences, and iterate. We make sure that in each case the squared number exceeds the corresponding number in the sequence going sequentially, so that when the max is taken one gets only squares; we also take care that we do not repeat squares, so that the resulting sequence of max terms is a subsequence of the sequence $1^2,2^2,3^2,...$, which ends up making the sum of max reciprocals converge.

I'll call $A$ the list of $a_k$ and $B$ the list of $b_k$. We will begin $A$ with a sequential part; since $1/1+1/2=1.5$ we may begin $a$ by $[1,2]$. For corresponding squares in $B$ we begin $B$ by $[1^2,2^2]$. We now want to switch to making adjacent integers in $B$. So we choose the least integer exceeding the present last term of $B$, which is 5. We find that $1/5+1/6+...+1/20$ is about 1.51 (stopping at 1/19 was too small]. So now we have $B=[1^2,2^2,5,6,...,20]$ The first square we haven't yet used is $3^2$, and (luckily) $3^2>5$, so we may now put $3^2$ into $A$ in its third position, and continue with $4^2,5^2,...,18^2$. At this point we have "done two blocks" of each sequence for $A$ and $B$, so that the beginnings of these are $A=[1,2,3^2,4^2,...,18^2],$ $B=[1^2,2^2,5,6,7,...,20].$ Now $18^2=324$, so the next entry of $A$ should be 325, in order to maintain an increasing sequence. And we may begin the squares in the next part of $B$ with $19^2$, since it happens to exceed 325. (In such cases, if the next square not yet used is not large enough, we may simply choose a larger square). For $A$ our next sequence is to be adjacent, and achieve the divergence bound of 1.5; I found we had to go all the way from 325 to 1453 for this. So the next block of adjacent $A$ values is quite long. The corresponding squares to be placed under these in $B$ may be taken to be $19^2,20^2,...,1147^2$.

Note: at the junctures I chose a square to place in one of the sequences, at the beginning of the part of the other which was to run in adjacent increasing form. The first square was say $a^2$ and at that term in the other was say $b$. I did make sure that $a^2>b$, but didn't explicitly say that the sequence of squares starting at $a^2$ automatically should exceed term by term the numbers following $b$. However this is clear, since the differences between succesive squares is always greater than 1, while the other sequence is only going up 1 each time.

It seems clear this construction will give what we want, since the partial sums of reciprocals for both of $A,B$ will exceed arbitrary multiples of the divergence bound of 1.5, making these series diverge, and yet the series of reciprocal max terms is bounded above by the sum of reciprocals of squares.

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That's not monotonic. – Thomas Andrews Oct 30 '12 at 23:26
Marvis, Thomas Andrews: Please have a look at the above, in which I definitely have (strictly) monotonic $a_n$ and $b_n$. I think it works now... – coffeemath Oct 31 '12 at 3:39
@coffeemath Thanks. I will look at it and get back to you. – user17762 Oct 31 '12 at 3:43
@coffeemath Thanks for the answer. Yes this does the job. +1. I have accepted did's answer since I can accept only one. – user17762 Jan 14 '13 at 4:50