# $\sum_{n=1}^{\infty} \frac{a_n}{b_n}$ converges?

If $$\sum_{n=1}^{\infty}a_n$$ converges, $$\sum_{n=1}^{\infty}b_n$$ diverges, but $$b_n \not=0 \forall n \in \mathbb{N}$$, then $$\sum_{n=1}^{\infty} \frac{a_n}{b_n}$$ converges? No, I think it's not true. Let $$a_n=\frac{1}{n^2}$$, $$b_n = \frac{1}{n}$$. Then it diverges.

If $$\lim_{n \to \infty}a_n=0$$ and $$\sum_{n=1}^{\infty}b_n$$ converges, then $$\sum_{n=1}^{\infty}a_nb_n$$ converges? No, I think it's not true. $$a_n=n$$, $$b_n = \frac{1}{n^2}$$, so $$\sum a_nb_n$$ diverges.

• Your counterexamples are not good. Apr 20, 2021 at 21:14
• Try proving that $\sum 1/n^p$ converges for all $p>1.$ Apr 20, 2021 at 21:17
• Do you assume the sequences are positive? Because if not, take $b_n = (-1)^n$ and $a_n = \frac{(-1)^n}{n}$. Apr 20, 2021 at 21:28
• @oraora Much better Apr 20, 2021 at 21:39
• @oraora First counterexample is correct. Second one is wrong: $\lim_{n\to\infty}n\neq0$. Apr 20, 2021 at 21:41

## 1 Answer

For the second, take

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

then

$$\lim_{n\to\infty}a_n=0$$

$$\sum b_n \text{ alternate convergent}$$ $$\sum a_nb_n=\sum \frac 1n\text{ divergent}$$