Simplifying $\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\min(n,m)}a_{l}b_{m-l}c_{n-l}$ I have come across a sum of the following form;
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\min(n,m)}a_{l}b_{m-l}c_{n-l}$$
and want to simplify it (in particular to remove the $min(n,m)$). I believe (although am not 100% sure), that it can be reduced simply to;
$$\left(\sum_{i=0}^{\infty}a_{i}\right)\left(\sum_{j=0}^{\infty}b_{j}\right)\left(\sum_{k=0}^{\infty}c_{k}\right)$$
I have been trying to get it into the form of the Cauchy product, which for three sums reads;
$$
\left(\sum_{i=0}^{\infty}a_{i}\right)\left(\sum_{j=0}^{\infty}b_{j}\right)\left(\sum_{k=0}^{\infty}c_{k}\right)
=
\sum_{k_{1}=0}^{\infty}\sum_{k_{2}=0}^{k_{1}}\sum_{k_{3}=0}^{k_{2}}a_{k_{1}-k_{2}}b_{k_{2}-k_{3}}c_{k_{3}}
$$
but haven't had any luck.
To deal with the $\min(n,m)$ term I have tried splitting up the sum into three parts; $n<m$, $m<n$, and $n=m$, yielding;
$$
\sum_{n=0}^{\infty}\sum_{m=0}^{n}\sum_{l=0}^{m-1}a_{l}b_{m-l}c_{n-l}
+
\sum_{m=0}^{\infty}\sum_{n=0}^{m}\sum_{l=0}^{n-1}a_{l}b_{m-l}c_{n-l}
+
\sum_{n=0}^{\infty}\sum_{l=0}^{n}a_{l}b_{m-l}c_{n-l}
$$
Which is close to the Cauchy product form, but unfortunately not close enough. Any help on this would be much appreciated.
 A: Write your sum in this way
$$
\eqalign{
  & \sum\limits_{n = 0}^\infty  {\sum\limits_{m = 0}^\infty  {\sum\limits_{l = 0}^{\min \left( {n,m} \right)} {a_{\,l} \;b_{\,m - l} \;c_{\,n - l} } } }  =   \cr 
  &  = \sum\limits_{\left( {l,m,n} \right)\; \in \;C} {a_{\,l} \;b_{\,m - l} \;c_{\,n - l} } \quad \left| {\;C = \left\{ {\left( {l,m,n} \right)} \right\}:\left\{ \matrix{
  0 \le l \le \min (n,m) \hfill \cr 
  0 \le m \hfill \cr 
  0 \le n \hfill \cr}  \right.} \right. \cr} 
$$
Then
$$
\eqalign{
  & \left\{ \matrix{
  0 \le l \le \min (n,m) \hfill \cr   0 \le m \hfill \cr 0 \le n \hfill \cr}  \right.
   = \left\{ \matrix{
  0 \le m < n \hfill \cr   0 \le l \le m \hfill \cr}  \right.\;
 \cup \;\left\{ \matrix{  0 \le n \le m \hfill \cr   0 \le l \le n \hfill \cr}  \right.
 =   \cr 
  &  = \left\{ {0 \le l \le m < n} \right\}\; \cup \;\left\{ {0 \le l \le n \le m} \right\} =   \cr 
  &  = \left\{ \matrix{
  0 \le l \hfill \cr   0 \le m - l < n - l \hfill \cr   1 \le n - l \hfill \cr}  \right.\;\;
 \cup \;\;\left\{ \matrix{
  0 \le l \hfill \cr   0 \le n - l \le m - l \hfill \cr   0 \le m - l \hfill \cr}  \right. \cr} 
$$
and the two sets are cearly disjoint.
Thereafter
$$
\eqalign{
  & \sum\limits_{\left( {l,m,n} \right)\; \in \;C} {a_{\,l} \;b_{\,m - l} \;c_{\,n - l} }
  = \sum\limits_{\left( {l,m,n} \right)\; \in \;C_{\,1} } {a_{\,l} \;b_{\,m - l} \;c_{\,n - l} }  + \sum\limits_{\left( {l,m,n} \right)\; \in \;C_{\,2} } {a_{\,l} \;b_{\,m - l} \;c_{\,n - l} }  =   \cr 
  &  = \sum\limits_{0\, \le \,l} {\;\sum\limits_{1\, \le \,k} {\;\sum\limits_{0\, \le \,j\, < \,k} {a_{\,l} \;b_{\,j} \;c_{\,k} } } }
  + \sum\limits_{0\, \le \,l} {\;\sum\limits_{0\, \le \,j} {\,\sum\limits_{0\, \le \,k\, \le \,j} {a_{\,l} \;b_{\,j} \;c_{\,k} } } }  =   \cr 
  &  = \left( {\sum\limits_{0\, \le \,l} {a_{\,l} \;} } \right)\left( {\sum\limits_{1\, \le \,k} {\;c_{\,k} \sum\limits_{0\, \le \,j\, < \,k} {\;b_{\,j} \;} }
  + \sum\limits_{0\, \le \,j} {\,b_{\,j} \sum\limits_{0\, \le \,k\, \le \,j} {\;\;c_{\,k} } } } \right) =   \cr 
  &  = \left( {\sum\limits_{0\, \le \,l} {a_{\,l} \;} } \right)\left( {\sum\limits_{0\, \le \,j\, < \,k} {b_{\,j} \;c_{\,k} }
  + \sum\limits_{0\, \le \,k\, \le \,j} {b_{\,j} \;c_{\,k} } } \right) =   \cr 
  &  = \left( {\sum\limits_{0\, \le \,l} {a_{\,l} \;} } \right)\left( {\sum\limits_{0\, \le \,j\,,\,\,k} {b_{\,j} \;c_{\,k} } } \right) =   \cr 
  &  = \left( {\sum\limits_{0\, \le \,l} {a_{\,l} \;} } \right)\left( {\sum\limits_{0\, \le \,j} {b_{\,j} } } \right)
\left( {\sum\limits_{0\, \le \,\,k} {c_{\,k} } } \right) \cr} 
$$
result that we could also obtain continuing to manipulate
the set of inequalities above.
A: Let $I = \{ (n, m, l) : 0 \leqslant l \leqslant \min\{n, m\} \}$,
$J = \mathbb{N}^3$, $f: I \to J$, $f(n, m, l) = (l, m-l, n-l)$, and
$g: J \to I$, $g(i, j, k) = (i + k, i + j, i)$. Then $f$ and $g$ are
mutually inverse bijections.
Let $(a_i), (b_j), (c_k)$ be absolutely convergent series of complex
numbers, with sums $A, B, C$. By two applications of proposition
(5.5.3) of Dieudonné, Foundations of Modern Analysis, the denumerable
family of complex numbers $(a_ib_jc_k)_{(i,j,k)\in J}$ is absolutely
summable, with sum $ABC$. Therefore the family of complex numbers
$(a_lb_{m-l}c_{n-l})_{(n,m,l)\in I}$ is absolutely summable, with
sum $ABC$.
$I$ is the disjoint union of the sets $K_n = \{ (n, m, l) : 0
\leqslant l \leqslant \min\{n, m\} \}$, and each $K_n$ is the disjoint
union of the sets $L_{n, m} = \{ (n, m, l) : 0 \leqslant l \leqslant
\min\{n, m\} \}$. By two applications of Dieudonné's proposition
(5.3.6):
$$
\sum_{n=0}^\infty \sum_{m=0}^\infty \sum_{l=0}^{\min(n,m)}
a_lb_{m-l}c_{n-l} = \!\!\!
\sum_{(n,m,l)\in I} a_lb_{m-l}c_{n-l} = ABC,
$$
and all the series on the left are absolutely convergent.
