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

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?

## 1 Answer

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 $\sum_{p=0}^{\infty} a_p \sum_{q=0}^{\infty} b_q$.