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If $a_n,b_n>0$ and $\sum{a_n}$ and $\sum{b_n}$ converge, then show whether $\sum{a_nb_n}$ converge or diverge where n={1,2,3...}

I can't think of a way to solve this. I tried solving it using their sequence of partial sum but realized that sequence of partial sum for the product of the two series is complicated.

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    $\begingroup$ If $c_n$ is any bounded sequence, does $\sum a_n c_n$ converge? $\endgroup$ – Daniel Fischer May 13 '17 at 11:25
  • $\begingroup$ @DanielFischer $c_n = (-1)^n$ and $a_n=1$...diverge $\endgroup$ – Vishweshwar Tyagi May 13 '17 at 11:34
  • $\begingroup$ We have the assumption that $\sum a_n$ converges. (And $a_n > 0$.) $\endgroup$ – Daniel Fischer May 13 '17 at 11:48
  • $\begingroup$ $\sum{a_n}$ converges then $a_n -> 0$ as $n->\infty$ then I think $\sum{a_nc_n}$ should converge by Alternating Series Test. But please confirm with someone else aswell $\endgroup$ – Vishweshwar Tyagi May 13 '17 at 12:15
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    $\begingroup$ The alternating series test isn't the thing here (generally, $(c_n)$ doesn't need to have alternating sign, nor does $\lvert a_n c_n\rvert$ need to be monotonic). The thing is absolute convergence. If $\sum x_n$ is absolutely convergent, and $(y_n)$ is bounded, then $\sum x_n y_n$ is absolutely convergent. For if $\lvert y_n\rvert \leqslant K$ for all $n$, then $\sum \lvert x_ny_n\rvert \leqslant K\sum \lvert x_n\rvert$. $\endgroup$ – Daniel Fischer May 13 '17 at 12:31
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You have that $\sum b_n$ converge, so $b_n$ must tend to $0$, hence, it's bounded, let's denote the upper bound $M$, Then, since $b_n$ is positive:

$\sum a_nb_n<M\sum a_n$

But $\sum a_n$ converges, so, since $a_n$ is also positive, we conclude that :

$\sum a_nb_n$ converges

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