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Let $\{a_n\}$ and $\{b_n\}$ be two decreasing sequences of positive numbers. Assume that both $$\sum_{n=1}^{\infty}a_n=\infty,\text{ }\sum_{n=1}^{\infty}b_n=\infty.$$

Let $I\subset\mathbb{N}$ and define $$c_n:=\begin{cases}a_n,&\text{ for }n\in I\\b_n,&\text{ for }n\notin I.\end{cases}$$

Can we have $$\sum_{n=1}^{\infty}c_n<\infty?$$

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The answer is yes, and the following is an example.

Let $S_{-1}=0$ and for $n\ge 0$, let $$S_n=\sum_{k=0}^n 2^{2^k}. $$ For every $n\ge 0$, let $$I_n=\{k\in\Bbb N: S_{n-1}<k\le S_n\}$$ and define $$a_k=2^{-2^{2n}-n}\quad\text{and}\quad b_k=2^{-2^{2n}},\quad\text{when}\quad k\in I_{2n};$$ $$a_k=2^{-2^{2n+1}}\quad\text{and}\quad b_k=2^{-2^{2n+1}-n},\quad\text{when}\quad k\in I_{2n+1}.$$ By definition, it is easy to verify that both $(a_n)$ and $(b_n)$ are decreasing. Let $$I=\bigcup_{n=0}^\infty I_{2n}.$$ Then it is easy to verify that for every $n\ge 0$, $$\sum_{k\in I_{2n}}a_k=\sum_{k\in I_{2n+1}}b_k=2^{-n}\quad\text{and}\quad\sum_{k\in I_{2n+1}}a_k=\sum_{k\in I_{2n}}b_k=1.$$ Therefore, $$\sum_{n\in I}a_n=2,\quad \sum_{n\notin I}a_n=\infty,\quad\sum_{n\in I}b_n=\infty,\quad\sum_{n\notin I}b_n=2.$$

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  • $\begingroup$ The phrase "it is easy to verify" is terrible. $\endgroup$ – Antonio Vargas Aug 2 '13 at 20:02
  • $\begingroup$ @AntonioVargas: Which one? Or both? $\endgroup$ – 23rd Aug 2 '13 at 20:02
  • $\begingroup$ Always, in my opinion. $\endgroup$ – Antonio Vargas Aug 2 '13 at 20:03
  • $\begingroup$ @AntonioVargas: Then you may just skip this phrase in this particular case. $\endgroup$ – 23rd Aug 2 '13 at 20:05
  • $\begingroup$ There is a misplaced $-$ sign in the definition of $b_k$ in the first group of chunks. How hard/easy did you find it? Would you put it in a calculus exam? I think having seen the proof of the divergence of $\sum 1/n$ should indicate them the idea. $\endgroup$ – OR. Aug 2 '13 at 20:07

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