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


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, $$\sum_{n\in I}a_n=2,\quad \sum_{n\notin I}a_n=\infty,\quad\sum_{n\in I}b_n=\infty,\quad\sum_{n\notin I}b_n=2.$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.