Simplifying $\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\min(n,m)}a_{l}b_{m-l}c_{n-l}$

Question

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.

Answer 1

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.

Answer 2

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.