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Do there exist sequences $\{a_n\}$ and $\{b_n\}$ satisfying all of the following properties?

  • $a_n>0$ and $b_n>0$
  • $\{a_n\}$ and $\{b_n\}$ are both decreasing
  • $\sum a_n$ and $\sum b_n$ both diverge
  • $\sum\min\{a_n,b_n\}$ converges
Brian Fitzpatrick
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Takanashi
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  • "can it" and "does it" are very different questions — "does it" usually implies that it always converges; "can it" just means find an example where it is true. – pancini Feb 09 '16 at 00:29
  • what if $\sum_n a_{2n}$ and $\sum_n b_{2n+1}$ both converge and $\sum_n a_{2n+1}$ and $\sum_n b_{2n}$ both diverge ? – reuns Feb 09 '16 at 00:34
  • @ElliotG Thank you, I will make my question clearer – Takanashi Feb 09 '16 at 00:36
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    Yes. What if you make $\min[a_n,b_n]$ as any decreasing sequence you like that has a convergent sum, say, $\min[a_n,b_n] = 2^{-n}$, and then oscillate in your definition of $a_n$ and $b_n$, making one relatively flat, then the other? So $a_n$ achieves the min for a while while $b_n$ essentially stays the same, then vice versa. – Michael Feb 09 '16 at 00:54
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    Anyway, if this is a homework problem, that should be a pretty good hint. – Michael Feb 09 '16 at 01:01
  • @Michael this is not a homework. Since I want {an} and {bn} to be decreasing, then how to oscillate them? – Takanashi Feb 09 '16 at 01:06
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    What is the context of the problem? – Michael Feb 09 '16 at 01:06
  • Note that you can replace the "decreasing" constraint with "non-increasing" to make the solution simpler, then just perturb the solution a bit to get both $a_n$ and $b_n$ decreasing. That is why I said "relatively flat" instead of "flat". – Michael Feb 09 '16 at 01:07
  • @Michael I know what you mean, but I not sure this will make {an} and {bn} diverge – Takanashi Feb 09 '16 at 01:12
  • Let us continue this discussion in chat. – Michael Feb 09 '16 at 01:12

1 Answers1

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Edit: Oops. Looking at the comments I see this is exactly what Michael has been suggesting. Sorry - these things happen.

Yes.

Say $1=N_1<N_2<\dots$. Define $a_n$ and $b_n$ like so: Assume $N_j\le n<N_{j+1}$.

If $j$ is odd set $a_n=1/n^2$, $b_n=1/N_j^2$.

If $j$ is even set $a_n=1/N_j^2$, $b_n=1/n^2$.

Then $\min(a_n,b_n)=1/n^2$. A little head-scratching shows that both sequences are decreasing. And it's clear that if we take each $N_{j+1}-N_j$ large enough then both $\sum a_n$ and $\sum b_n$ diverge (for example if $N_{j+1}-N_j>N_j^2$).

(I'm assuming that decreasing means non-increasing. You could easily jiggle the above a little to get strictly decreasing sequences.)

David C. Ullrich
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