I am thinking if I could get help for the following question:
Given a formal power series
$$g(z)=\sum_{i=0}^\infty a_i z^{-i}$$
does there always exists another (non-trivial) formal power series $y(z)$, such that the Cauchy product between $y$ and $g$
$$
y \times g= \sum_{i=0}^\infty c_k z^{-k}
$$
satisfies that
$$
\sum_{i=0}^\infty |c_k| < \infty?
$$
Best Answer
If the constant term $a_0$ of $g$ is nonzero, then $g$ will be invertible: that is, there is some $y$ such that $y \times g=1$.
If $g$ is nonzero (but we don't make any assumptions about its constant term), let $a_k$ be the first nonzero coefficient of $g$. We then have
\begin{align*} g(z)&=\sum_{n=0}^\infty a_nz^n\\ &=\sum_{n=k}^\infty a_nz^n&&\text{(because all the prior terms are zero)}\\ &=z^k\sum_{n=k}^\infty a_nz^{n-k}&&\text{(factoring out the common factor of }z^k\text{ from each term)}\\ &=z^k\sum_{n=0}^\infty a_{n+k}z^n&&\text{(relabeling).} \end{align*}
The sum in the last line is a power series with nonzero constant term $a_k$. That is, we can write $g=z^kh$, where $h$ is a power series with nonzero constant term. So if we take $y$ to be the multiplicative inverse of $h$, then $y \times g=z^k$.
Finally, if $g=0$, we can take $y$ to be anything we want and have $y \times g=0$.
So, for any power series $g$, we can find $y$ such that $y \times g$ such that finitely many of the $c_k$ (in fact, at most one of the $c_k$!) are nonzero, which means that the sum of their absolute values must converge.