1

The only approach I can think of is using the fact that $\{a_n\}$ and $\{b_n\}$ are both bounded as they are convergent then applying it to $\sum a_n b_n$ and saying it is bounded and increasing/dec and monotonic. However, I don't think we are allowed to prove using the Cauchy product and I'm unsure how to go about this approach (if it is even right)

Andres Mejia
  • 20,977
Joemans
  • 115
  • 1
    hint: Boundness of $b_n$ with direct comparison test. It has nothing to do with Cauchy product. – user251257 Mar 17 '16 at 01:47
  • lemma : if $|f(n+1)| \le |f(n)|$ and $\sum_n |a_n| < \infty$ then $\sum_n |f(n) a_n| < \infty$. now, given that the order of summation doesn't matter for absolutely convergent series, sort $b_n$ such that $b_{n+1} \le b_n$, and you are done – reuns Mar 17 '16 at 01:48

3 Answers3

7
  1. Since $\sum_{n=1}^\infty b_n$ converges, there exists such $N$ that $\forall n\geq{N}\,\,b_n\leq{1}$.
  2. Convergence of $\sum_{n=1}^\infty a_nb_n$ is equivalent to convergence of $\sum_{n=N}^\infty a_nb_n$.

Can you take it from here?

Vossler
  • 1,080
  • im confused, can you give me the answer – Joemans Mar 17 '16 at 01:54
  • 5
    @Joemans The policy of this site is not to provide complete solutions to problems like this (thus discouraging any motivation to learn and think for yourself), but to give advice to encourage the learning process. The hints that I (and the other respondents) gave you are more than enough to finish the job. – Vossler Mar 17 '16 at 02:05
  • @Vossler I want to say +1 for sticking to your guns, but really I just like this hint better than mine. – Andres Mejia Mar 17 '16 at 02:08
  • 1
    @AndresMejia cheers! Funnily enough, I, on the other hand, like your hint more... – Vossler Mar 17 '16 at 02:30
  • @Vossler It looks nicer, but yours makes sense really of the proof for my hint (or at least gets you going in the right direction): http://math.stackexchange.com/questions/202467/cauchy-product-of-two-absolutely-convergent-series-is-absolutely-convergent-ru – Andres Mejia Mar 17 '16 at 02:31
7

hint: $\sum a_n b_n \leq (\sum a_n)(\sum b_n)$

since $a_n \geq 0$ and $b_n \geq 0$ for all $n \in \mathbb{N}$, and since both series converge, we have boundedness, so we are almost done!

Andres Mejia
  • 20,977
  • thank you! I've seen you before: http://math.stackexchange.com/questions/1611560/why-are-all-k-cells-convex/1611571#1611571. – Andres Mejia Mar 17 '16 at 02:14
1

For all sufficiently large $n$ we have $0<b_n<1$ so for all sufficiently large $n$ we have $0<a_nb_n<a_n$ so $$\lim_{n\to \infty}\sup_{m\geq 0}|\sum_{j=0}^{j=m}a_{n+j}b_{n+j}|\leq \lim_{n\to \infty}\sup_{m\geq 0}|\sum_{j=0}^{j=m}a_n|=0.$$ With some appropriate changes from $<$ to $\leq$ we see that this also holds if $a_n$ and $b_n$ are non-negative.

DanielWainfleet
  • 57,985