Hello,


working on some machine learning problem I end up facing a problem which looks like generalizing the notion of [Cauchy product](http://en.wikipedia.org/wiki/Cauchy_product).


I briefly go back to Cauchy products before exposing my question. Consider, two sequences $(a\_n)\_{n \in \mathbb N}$ and $(b\_n)\_{n \in \mathbb N}$, which are assumed to be absolutely convergent (for simplicity). Then one can define another sequence $c\_n = \sum\_{k = 0}^{n}a\_k b\_{n-k}$ such that
$$\sum\_{n=0}^{+\infty} c\_n = \Big(\sum\_{n=0}^{+\infty} a\_n\Big)\Big(\sum\_{n=0}^{+\infty} b\_n\Big)$$
In the simpler framework, this is known as Mertens theorem and the (pedestrian) proof can be found in [this wikipedia page](http://en.wikipedia.org/wiki/Cauchy_product#Proof_of_Mertens.27_theorem).


My question consists in the possible generalization / extension of such result to the case where the sequence $b$ would have two indices. More precisely, if one introduces another sequence $(\theta\_n)\_{n \in \mathbb N}$, I would like to consider the case of
$$b\_k^n = \prod\_{i = k}^{n} \theta\_i$$
and I am interested in computing the double sum
$$\sum\_{n=0}^{+\infty}\sum\_{k = 0}^{n}a\_k b\_k^n$$
Note that, if all the $\theta\_i$ are equal (and smaller than 1), then we fall under Mertens theorem range of application. But what if the $\theta\_i$ are different? Is it still possible to have such a result where the sums eventually separate?


I have tried but failed to extend Mertens theorem's proof to this case. Any link to references which may help me figure this out would be most welcome. Besides, if one could show me a path to learn more about sequence with two indices, I would be very happy!


Cheers



Generalization of Cauchy product
If the $\theta\_i$ are different (but still all $|\theta\_i| < 1 - \epsilon < 1$ for some fixed $\epsilon$), then 
$$ \sum\_{n=0}^\infty \sum\_{k=0}^n \left|a\_k \prod\_{i=k}^n \theta\_i \right| 
< \sum\_{k=0}^\infty \sum\_{n=k}^\infty |a\_k| (1-\epsilon)^{n-k+1} = \sum\_{k=0}^\infty |a\_k| \frac{1-\epsilon}{\epsilon}$$
so it converges absolutely if $\sum\_{k=0}^\infty |a\_k|$ converges absolutely, and then we can rearrange:
$$ \sum\_{n=0}^\infty \sum\_{k=0}^n a\_k \prod\_{i=k}^n \theta\_i = \sum\_{k=0}^\infty \sum\_{n=k}^\infty a\_k \prod\_{i=k}^n \theta\_i = \sum\_{k=0}^\infty a\_k c\_k$$
where $c\_k = \sum\_{n=k}^\infty \prod\_{i=k}^n \theta\_i = \theta\_k (1 + \theta\_{k+1} (1 + \theta\_{k+2} (\ldots)))$. Thus $|c\_k| < (1 - \epsilon)/\epsilon$ and $c\_k = \theta\_k (1 + c\_{k+1})$. 


