TITLE: Does the minimum of two decreasing divergent series diverge?
QUESTION [9 upvotes]: 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..)

REPLY [3 votes]: 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 $B_n$.)