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Assume that $\sum a_n$ converges, $a_n \in \mathbb{R}$, then there exists real sequence $b_n$ such that $b_n\rightarrow +\infty$ and $\sum a_n b_n$ converges.

Same to be easy at first thought, can we find such $b_n$ represented by $a_n$?

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  • $\begingroup$ I'm not sure what you mean by "can we find $b_n$ such represented by $a_n$?" Are you asking if we can use the $a_n$'s to construct $b_n$'s? Also, what approaches have you tried so far? $\endgroup$ – vociferous_rutabaga Jul 24 '14 at 15:29
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    $\begingroup$ See this. $\endgroup$ – David Mitra Jul 24 '14 at 15:59
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Credit: Rudin's "Principles of Mathematical Analysis", Chapter 3, Exercise 12.

Let $$r_n=\sum^\infty_{m=n}a_n.$$ Then $\sum\frac{a_n}{\sqrt{r_n}}$ is convergent.

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    $\begingroup$ How do you deal with $r_n = 0$ for some (say infinitely many) $n$? $\endgroup$ – Daniel Fischer Jul 24 '14 at 15:38
  • $\begingroup$ Apparently, the exercise is only for $a_n>0$. One can partition a general series to get a positive/negative series if needed, similar to @nayrb 's answer. $\endgroup$ – Quang Hoang Jul 24 '14 at 15:45
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HINT: Because $\sum a_n$ converges to $A$, we can partition it into sets

$$\{a_1, \dots, a_{N_1}\}, \{a_{N_1+1}, \dots, a_{N_2}\}, \dots$$

where $$\sum_{i={N_k+1}}^{N_{k+1}} a_i \approx \frac{A}{4^k}.$$

Then if you pick $b_n$ as the sequence $B_k$ but with a lot of repetition, you can make $\sum a_n b_n \approx \sum \frac{A}{4^k} B_k$.

What remains to be done is choosing the numbers $N_k$ precisely and making the approximate equality more precise by changing it to an inequality and/or modifying $A/4^k$. There might be a snag with alternating sums, that is sums that have negative terms, but I think the essence of the idea is here.

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