Let $ a_j, b_j \in \mathbb C$ for all $j$ such that both $\sum_{j=0}^{\infty}|a_n|$ ,$\sum_{j=0}^{\infty}|b_n|$ converges.

Show that $\sum_{n=0}^{\infty}(\sum_{j=0}^n a_jb_{n-j})$ converges to $(\sum_{j=0}^{\infty}b_n)(\sum_{j=0}^{\infty}a_n)$.

I know that since the two complex series converges absolutely, they converge, thus they're both bounded.

Let $|b_j|< K$ for all $j$ where $K>0$
there exist $J_1$ such that for all $j \ge J_1$, $|b_j -M| < \frac{\epsilon}{2L}$
there exist $J_2$ such that for all $j \ge J_2$, $|a_j -L| < \frac{\epsilon}{2K}$
then for all $j \ge J$ where $J = \max\{J_1,J_2\}$ \begin{align*} |a_jb_j - LM| &= |(a_jb_j - Lb_j) + (Lb_j - LM)|\\ &\le |a_j -L||b_j| + |L||b_j - M|\\ &\le |a_j -L|K+L|b_j-M|\\ &\le \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Would this be the right way to complete this question?


Note that $\displaystyle \sum_{p=0}^{\lfloor N/2 \rfloor} a_p \sum_{q=0}^{N-\lfloor N/2 \rfloor} b_q = \sum_{n=0}^{N} \sum_{p+q=n} a_pb_q = \sum_{n=0}^N\sum_{p=0}^na_pb_{n-p}$

Since the sequences $N-\lfloor N/2 \rfloor$ and $\lfloor N/2 \rfloor$ both go to $\infty$ monotonically, $\displaystyle \lim_{N\to \infty} \sum_{p=0}^{\lfloor N/2 \rfloor} a_p \sum_{q=0}^{N-\lfloor N/2 \rfloor} b_q =\sum_{p=0}^{\infty} a_p \sum_{q=0}^{\infty} b_q$

Therefore, the series $\sum_{n=0}^N\sum_{p=0}^na_pb_{n-p}$ is convergent and its limit is $\sum_{p=0}^{\infty} a_p \sum_{q=0}^{\infty} b_q$.

| cite | improve this answer | |

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.