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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.

Tundoku
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    Your title has two words and a typo! – Mariano Suárez-Álvarez Mar 28 '17 at 05:38
  • I think $\sum \min(a_n,b_n)$ must diverge. $\sum \min(a_n,b_n)$ is a sum over infinitely many terms from $a_n$ and the complement of those terms from $b_n$. One of these infinite subsets will have positive upper density, so the sum over that subset should also diverge. I'm not sure if this is true though.. – mathworker21 Mar 28 '17 at 05:47
  • @Mariano Suárez-Álvarez I'm so sorry for my carelessness. I've corrected it. – Tundoku Mar 28 '17 at 05:47
  • @mathworker21 But it seems that without monotonicity we can give two such sequences. – Tundoku Mar 28 '17 at 06:33
  • Yes, the monotonicity will be useful for showing the sum over the subset with positive upper density will diverge. – mathworker21 Mar 28 '17 at 08:34

2 Answers2

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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:

  1. 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 $1\over(2n+2)!$.
  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 $1\over(2n+2)!$.
  3. Increment $n$ and repeat.
Ivan Neretin
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  • Same idea as I had, different execution. Basically, you can take any converging series with positive, deceasing elements as $\min(a_n,b_n)$, then take very long intervals, where $a_n$ is the same as $\min(a_n,b_n)$ and $b_n$ is constant, then exchange the role for the next very long interval. In those very long intervals, one series accumulates lots of value during summing, such that the overall sum diverges. – Ingix Mar 28 '17 at 09:12
  • Indeed. A similar idea is applicable to the question of converging $\sum\sqrt{a_nb_n}$, given the same premises. – Ivan Neretin Mar 28 '17 at 12:51
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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 $\sum b_n$.

Ingix
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