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 1


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 $$ \sum_{n=1}^\infty |a_n b_n| \le C\sum_{n=1}^\infty |a_n|<\infty.$$


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