1
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.