$\sum |a_n|<\infty$ and $|\sum b_n|<\infty$ implies $|\sum a_n b_n| <\infty$

Suppose $$\sum_{n} a_n$$ converges absolutely and $$\sum_{n} b_n$$ is any convergent series. Then $$\sum_n a_nb_n$$ is convergent.

Proof: Since $$\sum_{n} b_n$$ is convergent, we can choose $$N_0$$ s.t $$\forall n\geq N_0$$ we have $$|b_n|\leq 1/2$$. Similarly, since $$\sum_{n} a_n$$ converges absolutely, we can choose $$N'$$ s.t $$\forall m,n\geq N'$$ we have $$|a_n|+...+|a_m|\leq \epsilon$$. Therefore,

$$\forall \epsilon>0 \exists N=\max\{N_0,N'\}$$

such that

$$\forall m,n\geq N \quad |a_nb_n|+...+|a_mb_m|\leq 1/2(|a_n|+...+|a_m|)\leq \epsilon /2.$$

Hence $$\sum_n a_nb_n$$ is convergent by Cauchy's criterion.

Is the proof correct?

1 Answer

You could make your strategy a little more clear, but essentially this is correct.

Still, it is a little over-complicated. With your choice of $$N_0$$ you have

$$\sum_{n=1}^\infty |a_n b_n| \le \sum_{n=1}^{N_0-1}|a_n b_n| + \frac12\sum_{n=N_0}^{\infty}|a_n|<\infty,$$

which is all you need. This even shows absolute convergence.

By the way: The assumption that $$\sum b_n$$ converges is much too strong. The statement is true (with the same arguments) if the sequence $$(b_n)$$ is merely bounded. Indeed, $$|b_n|\le C$$ for all $$n\in \Bbb N$$ yields $$\sum_{n=1}^\infty |a_n b_n| \le C\sum_{n=1}^\infty |a_n|<\infty.$$

• oh im considering the cauchy criterion for convergence en.wikipedia.org/wiki/Cauchy%27s_convergence_test Mar 14, 2019 at 15:06
• Yes, sorry. I misread part of your proof. Mar 14, 2019 at 15:09