There are also infinite dimensional generalizations (say for bounded operators on Banach spaces) of the Jordan decomposition, but the situation is of course more complicated. I'm certainly not well informed on the state of the art. But with a sufficiently strong hypothesis, like compactness, one can achieve a pretty complete statement. The following result is already in §80 of Riesz & Nagy's classic _Functional Analysis_ (Ungar, 1955):

> For a compact operator $K$ on a Hilbert space, the spectrum is discrete (accumulating at $0$) and the invariant subspace corresponding to each eigenvalue is finite dimensional (hence, $K$ restricted to each such invariant subspace has the usual Jordan decomposition).

Another reference for a discussion of this kind is in Chapter VII of Dunford & Schwartz's _Linear Operators, Part I: General Theory_ (Interscience, 1958).

Results for more general bounded or unbounded operators are obstructed by the related [Invariant Subspace Problem](https://en.wikipedia.org/wiki/Invariant_subspace_problem), which is still open. It asks whether any bounded operator on a Banach space has a non-trivial closed invariant subspace. If such an operator without a non-trivial closed invariant subspace existed, it would behave very differently from any finite dimensional operator (of size larger than $1\times 1$), which is guaranteed to have a $1$-dimensional invariant subspace in each Jordan block. Even under more restrictive hypotheses, the examples of shift operators seem to confound reasonable notions of a _nilpotent operator_ such that a reasonable class of operators could always be decomposed as $S + N$, where $S$ is semi-simple (diagonalizable, essentially) and $N$ nilpotent. See for instance the discussion on this [math.SE question](http://math.stackexchange.com/q/907376/84080).