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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.

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

1 Answer 1

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For the second, take

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

then

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

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

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