What is Absolute convergence? Take: 
$$
(u*v)(k) = \sum_{i=-\infty}^\infty u(i)v(k-i).
$$
$k$ is there, it's because you want to define
$$
\ldots\ldots, (u*v)(-3), (u*v)(-2), (u*v)(-1), (u*v)(0), (u*v)(1), (u*v)(2), (u*v)(3), \ldots\ldots
$$
etc.  The number in the parentheses is $k$.  Thus, for example, when $k=4$, we have
\begin{align}
(u*v)(4) = \sum_{i=-\infty}^\infty u(i)v(4-i)
\end{align}
$$
= \cdots\cdots+u(-3)v(7)+u(-2)v(6)+u(-1)v(5)+u(0)v(4)+u(1)v(3)+u(2)v(2)
$$
$$
\phantom{={}} {}+u(3)v(1)+ u(4)v(0) + u(5) v(-1) + u(6)v(-2)+u(7)v(-3)+u(8)v(-4)+\cdots\cdots.
$$
Take:
$u(x)=x-2$
$v(x)=3x^2+x$
is this formula Absolute convergent?
if not, could someone give a clear and well written example with the correct u(x) and v(x)?
 A: As in the previous post, I want to emphasize again that the notation used for sequences in this question is functional, and so the terms are functions on $\Bbb Z$, and the following polynomials are not being interpreted as functions, of their indeterminate $x$, but rather as functions of the degree of each $x$, producing a sequence on $\Bbb Z$ with their coefficients. This seems to be the source of your difficulty.
The polynomial $u(x)=x-2$ represents the following sequence on $\Bbb Z$:
$\{\dots ,u(-1),u(0),u(1),u(2),\dots\}$
$\{\dots 0,-2,1,0,\dots\}$
Where the sequence elements not displayed are zero.
Similarly for $v(x)=3x^2+x$:
$\{\dots v(-1),v(0),v(1),v(2),v(3),\dots\}$
$\{\dots ,0,0,1,3,0,\dots\}$
The convolution produces a new sequence from these two sequences.
$(u*v)(k) = \sum_{i=-\infty}^\infty u(i)v(k-i)$
For $k=0$, $(u*v)(0) = \sum_{i=-\infty}^\infty u(i)v(-i)$. But look: when $i$ is positive, $v(-i)$ is always 0, and when $i$ is negative, $u(i)$ is always zero. So, the computation yields that $(u\ast v)(0)=0$. In fact, you will find that is the case for all $k\leq 0$.
For $k=4$, $(u*v)(4) = \sum_{i=-\infty}^\infty u(i)v(4-i)$. Since for $i\notin\{0,1\}$ the term $u(i)=0$, this sum reduces to $\sum_{i=0}^1 u(i)v(4-i)$, but again, $v(4-i)$ is zero for $\{0,1\}$. So, $(u\ast v)(4)=0$, and in fact $k\geq 4$ behaves the same way.
So far, the convolution has been zero for $k\leq 0$ and $k\geq 4$. We now look at what goes on for $k=1,2,3$. I will trim the infinite sums as I did in the last paragraph without describing why.
$(u*v)(1) = \sum_{i=-\infty}^\infty u(i)v(1-i)=\sum_{i=0}^1 u(i)v(1-i)=u(0)v(1)+u(1)v(0)=(-2)1+(1)0=-2$
$(u*v)(2) = \sum_{i=-\infty}^\infty u(i)v(2-i)=\sum_{i=0}^1 u(i)v(2-i)=u(0)v(2)+u(1)v(1)=(-2)3+(1)1=-5$
$(u*v)(3) = \sum_{i=-\infty}^\infty u(i)v(3-i)=\sum_{i=0}^1 u(i)v(2-i)=u(0)v(3)+u(1)v(2)=(-2)0+(1)3=3$.
What polynomial does this sequence describe? Well, it is nonzero outside of degrees $k=1,2,3$, and lining those coefficients up with their power of $x$ you get this:
$-2x-5x^2+3x^3$ Lo and behold, that is the product of the original two polynomials!
In general it can be shown that convolution of sequences on the integers matches polynomial multiplication (polynomials being nothing more than sequences with finitely many nonzero terms above $k=0$). If you know about Laurent polynomials (which would simply be sequence with finitely many nonzero terms) then you will find the convolution matches the polynomial multiplication in the ring of Laurent polynomials as well.
