If $\sum a_n$and$\sum b_n$diverge, can$\sum \min\{a_n,b_n\}$converge? [duplicate]

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Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges (3 answers)

Closed 9 years ago.

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

• pancini: "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.
• reuns: 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 ?
• Michael: 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},ドル and then oscillate in your definition of $a_n$ and $b_n,ドル 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: Anyway, if this is a homework problem, that should be a pretty good hint.
• Michael: What is the context of the problem?

Answer 1

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

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Yes.

Say 1ドル=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,ドル $b_n=1/N_j^2$.

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

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.)