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Let $\sum_{i=0}^\infty a_n$ be a conditionally convergent series, and $\{b_n\}$ be a nonnegative and convergent sequence of real or complex numbers. Does $\sum_{i=0}^\infty a_n b_n$ converge?

Do we actually need convergence of $\{b_n\}$ for convergence of $\sum_{i=0}^\infty a_n b_n$ or is it sufficient that $\{b_n\}$ is nonnegative and bounded?

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Consider $a_n = \frac{(-1)^n}{\sqrt{n}}$ and $b_n = 2018+\frac{(-1)^n}{\sqrt{n}}$. Then all the conditions are met, although we have

$$\sum_{n=1}^{\infty} a_n b_n = \sum_{n=1}^{\infty} \left( 2018 \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}\right), $$

which diverges.

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  • $\begingroup$ Look good, thanks! $\endgroup$ – Solicitous Wookiee Dec 14 '18 at 16:54
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Bounded and non-negative is not sufficient. Consider $a_n=\frac{(-1)^n}n$ and $b_n=1+(-1)^n$.

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  • $\begingroup$ $b_n$ is not convergent $\endgroup$ – gimusi Dec 14 '18 at 16:52
  • $\begingroup$ That is the point... $\endgroup$ – SmileyCraft Dec 14 '18 at 16:53
  • $\begingroup$ @SmileyCraft The question assumes that $b_n$ is a convergent sequence. $\endgroup$ – BigbearZzz Dec 14 '18 at 16:56
  • $\begingroup$ The OP literally asks "is it sufficient that $\{b_n\}$ is nonnegative and bounded?" and my example answers this question. $\endgroup$ – SmileyCraft Dec 14 '18 at 16:59
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Assume

$$a_n = \frac{(-1)^n}{\sqrt n}$$

$$b_n =\begin{cases}=0\quad n\,\text{odd}\\\\=\frac{1}{\sqrt n}\quad n\,\text{even} \end{cases}$$

and therefore

$$\sum_{n=1}^{2N} a_n b_n=\sum_{n=1}^{N} \frac1{2n} \to\infty$$

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  • $\begingroup$ Your sequence $\;b_n\;$ isn't convergent... $\endgroup$ – DonAntonio Dec 14 '18 at 16:44
  • $\begingroup$ Opsss...thanks I fix $\endgroup$ – gimusi Dec 14 '18 at 16:44

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