Suppose that the sequence $\{a_n\}$ is unbounded. Prove that for some absolutely convergent series $\{b_n\}$, the series $\{a_nb_n\}$ is not absolutely convergent.

Absolute convergence means we must choose $\{b_n\}$ such that series $|b_n|$ converges, but series $|a_nb_n|$ does not converge.

I thought about choosing $b_n = 1/a_n$ which clearly makes $\{a_nb_n\}$ not absolutely convergent, but then I realized that even though $\{a_n\}$ is unbounded, it doesn't mean the series $|b_n|$ converges (for example, $\{a_n\} = 1,2,1,3,1,4,1,5,\ldots$.)

EDIT: Okay I might have gotten it. Since $\{a_n\}$ is unbounded, for any $k$ we can find $|a_{i_k}|>2^k$. So we have the subsequence $a_{i_1},a_{i_2},\ldots$. Choose $b_{i_k} = 1/a_{i_k}$ for all $k$, and $b_i=0$ otherwise. Then the series $|b_n|$ converges by comparison test with the geometric series $\{1/2^k\}$. On the other hand, the series $|a_nb_n|$ has infinitely many terms equal to $1$, implying that it diverges.


Because $\{a_n\}$ is unbounded, for any $k$ there is an $n$ such that $|a_n|\geq 2^k$. Define $$b_n=\begin{cases}\frac{1}{|a_n|} & \text{if there exists some }k\text{ such that }n\text{ is the first index such that }|a_n|\geq 2^k,\\[0.2in] \;\;0 & \text{otherwise} \end{cases}$$ Thus $$\begin{align*} \sum_{n=1}^\infty|b_n|&=\frac{1}{|a_1|}+0+\cdots+0+\frac{1}{|a_2|}+\cdots\\\\\\ &\leq\frac{1}{2^{k_0}}+0+\cdots+0+\frac{1}{2^{k_1}}+\cdots\\\\\\ &\leq\sum_{n=1}^\infty\frac{1}{2^n}=1 \end{align*}$$ but $$\sum_{n=1}^\infty |a_nb_n|=1+0+\cdots+0+1+\cdots=\infty.$$

  • $\begingroup$ Wow, exact same solution. You just beat me to it. Thank you Zev Chonoles :=) $\endgroup$ – PJ Miller Jun 15 '13 at 19:03

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