TITLE: why haven't certain well-researched classes of mathematical object been framed by category theory?
QUESTION [22 upvotes]: Category theory is doing/has done a stellar job on Set, FinSet, Grp, Cob, Vect, cartesian closed categories provide a setting for $\lambda$-calculus, and Baez wrote a paper (Physics, Topology, Logic and Computation: A Rosetta Stone) with Mike Stay about many of the interconnections between them. 
But there are mathematical objects that aren't thought of in a category-theoretic fashion, at least the extant literature doesn't tend to treat them as such. For instance nobody talks about Series, Products, IndefInt as being categories in their own right. (infinite series, infinite products, and indefinite integrals, respectively). (google searches for the phrase "the category of infinite series" in both the web and book databases have no hits whatsoever).  I suppose my question is: why not?

REPLY [3 votes]: I really like to think of $\mathbb{N}$-graded $R$-modules as power series $\bigoplus_{n \in \mathbb{N}} M_n \otimes X^{\otimes n}$, where each "coefficient" $M_n$ is an $R$-module and $X$ is the graded module concentrated degree $1$ and which is $R$ there. Therefore, we have a category of power series, where a morphism is just a family of morphisms between the coefficients. Actually this is a symmetric monoidal category - the tensor product is given by some convolution. And the same works if we replace $\mathsf{Mod}(R)$ by any cocomplete symmetric monoidal category. This is spelled out for example in Section 5.4 of my thesis.
In order to get a connection to power series in analysis, we might endow the unit interval $[0,1]$ with the structure of a cocomplete symmetric monoidal category (cf. Example 3.1.6 in loc.cit.): We use the usual ordering to make it a (thin) category and the usual multiplication to make it a symmetric monoidal category. Colimits are given by suprema. Therefore, we get a cocomplete symmetric monoidal category of sequences valued in $[0,1]$. This is again just an order with a multiplication, where we have $(a_n) \leq (b_n)$ iff $a_n \leq b_n$ for all $n$, and $((a_p) \cdot (b_q))_n = \sup_{p+q=n} a_p \cdot b_q$.
Also notice that coends in category theory capture some ideas of (definite) integration. See MO/78471 for some intuition.