# $\sum a_n b_n$ when $\sum a_n$ convergent and $\{b_n\}$ nonnegative

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?

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.

• Look good, thanks! – Solicitous Wookiee Dec 14 '18 at 16:54

Bounded and non-negative is not sufficient. Consider $$a_n=\frac{(-1)^n}n$$ and $$b_n=1+(-1)^n$$.

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

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

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

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