We have the Cauchy product $$ \sum_{k=0}^\infty a_k\sum_{\ell=0}^\infty b_\ell=\sum_{k=0}^\infty\sum_{\ell=0}^k a_\ell b_{k-\ell}. $$ Do we have a way of writing $$ \tag{1} \sum_{k=n}^\infty\sum_{\ell=0}^k a_\ell b_{k-\ell}=\sum_{k=?}^\infty a_?\sum_{\ell=?}^\infty b_?, $$ where $n=1,2,\dots$?
Certainly, $$ \begin{aligned} \sum_{k=n}^\infty\sum_{\ell=0}^k a_\ell b_{k-\ell} &=\sum_{k=n}^\infty\left(\sum_{\ell=0}^{n-1}+\sum_{\ell=n}^k\right) a_\ell b_{k-\ell}\\ &=\sum_{k=0}^\infty \sum_{\ell=0}^k a_{\ell+n} b_{k-\ell}+\sum_{\ell=0}^{n-1}a_\ell\sum_{k=0}^\infty b_{k+n-\ell}\\ &=\sum_{k=0}^\infty a_{k+n}\sum_{\ell=0}^\infty b_\ell+\sum_{\ell=0}^{n-1}a_\ell\sum_{k=0}^\infty b_{k+n-\ell}. \end{aligned} $$ However, I am curious if the "Cauchy product" in (1) can be written as a single product of two series without the additional sum $\sum_{\ell=0}^{n-1}$.
Edit: One thought I did have was to use differentiation and write something of the form $$ \partial_x^n\sum_{k=0}^\infty\sum_{\ell=0}^k a_\ell b_{k-\ell}x^k|_{x=1}. $$ Then you can factor out the product and differentiate. This of course doesn't exactly work but could be a potential lead.
You seek to factorize $\sum_{\ell+m\ge n}a_\ell b_m$. Since this includes an $a_0b_n$ term, your sum over $a$s needs to start at $a_0$. By the same logic, the sum over $b$s needs to start at $b_0$. So it's only possible if $n=0$.