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This question already has an answer here:

My attempt;

Since $a_n$ converges then its sequence of partial sums $(x_n)$ converges to a limit x. Also, since $b_n$ converges then its sequence of partial sums $(y_n)$ converges to a limit y. Observe that by using the multiplication limit theorem: If $x_n \to x$ and $y_n \to y$ then $x_n*y_n \to x*y$. We know that the product of its partials sums converge, thus the product of the series converges.$

Is this correct? Any guidance is appreciated !

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marked as duplicate by Martin R, JonMark Perry, Shailesh, Community Mar 13 '17 at 1:15

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    $\begingroup$ If $x_n=\sum_{k=1}^n a_k$ and $y_n=\sum_{k=1}^n b_k$ it is not true that $x_ny_n=\sum_{k=1}^{n}a_kb_k$. So the fact that $x_ny_n$ converges does help you. $\endgroup$ – Thomas Andrews Mar 11 '17 at 21:29
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    $\begingroup$ In particular, this theorem isn't true in general, although it is true if $\sum|a_n|$ and $\sum|b_n|$ converges. $\endgroup$ – Thomas Andrews Mar 11 '17 at 21:31
  • $\begingroup$ See here for the correct version of the theorem. $\endgroup$ – user228113 Mar 11 '17 at 21:35
  • $\begingroup$ If both sequences are non-negative, then you can argue that: because $\sum a_n$ converges, $a_n\rightarrow 0$; and so eventually $0 \le a_n \le 1$; and so eventually $0 \le a_n b_n \le b_n$; and so $\sum a_n b_n$ converges by comparison with $\sum b_n$. $\endgroup$ – mjqxxxx Mar 11 '17 at 22:17
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It is not generally true. For example, take $a_n=b_n=\frac{(-1)^n}{\sqrt n}$.

Clearly, $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ converge (Apply Leibniz's test for alternating series).

But, the series $\sum_{n=1}^\infty a_nb_n=\sum_{n=1}^\infty\frac1n$ diverges.

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This isn't true in general. there is a easy conterexample:

Regard $a_n =b_n= \frac{(-1)^n}{\sqrt{n}}$ then $a_nb_n = \frac{1}{n}$. you can use Leibniz criterion to check the convergence of $\sum^\infty_{n=1} a_n$ but $\sum^\infty_{n=1} a_nb_n = \sum^\infty_{n=1} \frac{1}{n}$ will not converge!

if you assume a little bit more you can fix that!

Either if one is $a_n$ or $b_n$ is monoton then you can dirichlets test or if one converge absolute then you can use holder inequality:

$\left|\sum_{n=1}^\infty a_n b_n\right| \leq \sup|a_n|\sum_{n=1}^\infty |b_n|$

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