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If $\displaystyle \sum_{n=1}^{\infty}a_n$ and $\displaystyle \sum_{n=1}^{\infty}b_n$ are both divergent series with $a_n\downarrow0$ and $b_n\downarrow0$,

[so $(a_n)$ and $(b_n)$ are decreasing sequences which converge to 0],

and if $c_n=\min\{a_n,b_n\},\;\;$ does the series $\displaystyle \sum_{n=1}^{\infty}c_n$ necessarily diverge?

(I was led to ask this question after reading this question and math110's solution to it: $a_n\downarrow 0, \sum\limits_{n=1}^{\infty}a_n=+\infty, b_n=min\{a_n,1/n\}$, prove $\sum b_n $ diverges..)

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user84413
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    If the individual terms tend to 0, they are not necessarily decreasing, as suggested by "so." Do you want them decreasing, or just tending to 0? – user217285 Jun 23 '15 at 18:42
  • @Nitin I wanted both, so I added the statement in brackets to clarify the notation I was using. – user84413 Jun 23 '15 at 18:53
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    My intuition tells me we can construct a counterexample out of slowly divergent series like the harmonic series, but I'm at a loss for how to actually do that. – Kevin Jun 23 '15 at 19:25
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    @AlexM. There isn't any full answer at your link, just hints. I'm not sure how this site handles that, i.e. if reposting of slightly modified unanswered questions is acceptable, since after enough time practically no one will find or revisit a posted question after a certain amount of time if it has up-voted answers. – user2566092 Jun 23 '15 at 19:53
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    I wonder about that as well, given that it's only hints there and I gave an actual answer... – David C. Ullrich Jun 23 '15 at 20:17

1 Answers1

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No, the minimum can converge. (Probably the minimum has to diverge if you add some convexity condition...)

First, the answer's clearly no if we omit the word "decreasing". Say $A_n,B_n>0$, $C_n=\min(A_n,B_n)$, $\sum A_n=\sum B_n=\infty$, $\sum C_n<\infty$.

Now, given a sequence of integers $0=k_0<k_1<k_2\dots$, define $$I_n=[k_n,k_{n+1}),$$where the interval notation is meant to refer to sets of positive intergers. Let $N_n$ be the number of elements of $I_n$.

Define $a_j,b_j$ and $c_j$ by $$a_j=A_n/N_n\quad(j\in I_n),$$ $$b_j=B_n/N_n\quad(j\in I_n),$$ and $$c_j=C_n/N_n\quad(j\in I_n).$$It's clear that $\sum_ja_j=\sum_nA_n=\infty$, and similarly for $b_j$ and $c_j$. And $c_j=\min(a_j,b_j)$. But if we choose the numbers $k_n$ appropriately then $a_j$ and $b_j$ are decreasing. (Choose them one at a time; having chosen $k_n$, choose $k_{n+1}$ so large that $A_{n}/N_{n}<A_{n-1}/N_{n-1}$, and similarly for $B_n$.)

David C. Ullrich
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