# If $\sum a_n$ and $\sum b_n$ and their convolution $\sum c_n$ converge, then $\sum c_n=\sum a_n\sum b_n$

Given $\sum a_n$ and $\sum b_n$ we define the Cauchy product to be $\sum c_n$ , where $c_n=\sum_{k = 0}^{n}a_kb_{n-k}$. It is well known that for $a_n,b_n \in \mathbb{R}$, if $\sum a_n$, $\sum b_n$ and $\sum c_n$ converges, then $\sum c_n=\sum a_n\sum b_n$. Is it true when $a_n,b_n \in \mathbb{C}$?

• Yes, it can be proved the same way. – Lord Shark the Unknown May 8 '18 at 3:33
• Typically: look at the proof, see if anything breaks. More likely than not, nothing will. – Clement C. May 8 '18 at 3:33
• @Lord Shark the Unknown Thanks for your answer! What is the same way? Is it to use abel’s theorem? – 04170706 May 8 '18 at 3:35
• Abel's theorem works fine @0706 – Lord Shark the Unknown May 8 '18 at 3:38
• How to solve in a more basic way? – ExactSequence Apr 15 at 6:51

Abel's Lemma. If $\{a_n\}\subset\mathbb R$ and $\sum a_n$ converges, then $\lim_{x\to 1^-}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_n$.
Clearly, Abel's Lemma holds even if $\{a_n\}\subset\mathbb C$, as we can write $a_n=\mu_n+i\nu_n$ and apply the lemma for the two real sequences $\{\mu_n\}$ and $\{\nu_n\}$.
Lemma. If the complex series $\sum a_n$ and $\sum b_n$ converge absolutely, then so does their convolution $c_n=a_0b_n+\cdots+a_nb_0$ and $\sum_{n=0}^\infty c_n=\sum_{n=0}^\infty a_n\sum_{n=0}^\infty b_n$.
In our case, since $\sum a_n$ and $\sum b_n$ converge, then the power series $\sum_{n=0}^\infty a_nz^n$ and $\sum_{n=0}^\infty b_nz^n$ have radius of convergence $R\ge 1$, and hence $\sum_{n=0}^\infty c_nz^n$ converges and $\sum_{n=0}^\infty c_nz^n=\Big(\sum_{n=0}^\infty a_nz^n\Big)\Big(\sum_{n=0}^\infty b_nz^n\Big)$, for all $|z|<1$.
Now, using Abel's Lemma (complex version) we have that $$\sum_{n=0}^\infty c_n=\lim_{x\to 1^-} \sum_{n=0}^\infty c_nx^n =\lim_{x\to 1^-}\big(\sum_{n=0}^\infty a_nx^n\Big)\big(\sum_{n=0}^\infty b_nx^n\Big)= \big(\sum_{n=0}^\infty a_n\Big)\big(\sum_{n=0}^\infty b_n\Big).$$