I'm wondering if a [monad](http://ncatlab.org/nlab/show/%28infinity%2C1%29-monad) $T$ on a stable $\infty$-category $\cal C$ has a stable $\infty$-category of algebras, provided $T$ preserves finite limits/colimits.


Is this true?


**Edit**: Is something similar true in the triangulated world?



Do the algebras for a $\infty$-monad form a stable $\infty$-category?
The statement about stable $\infty$-categories is true. Just as in ordinary categories the basic facts about limits and colimits in the category of algebras are that the forgetful functor $\mathrm{Alg}(T) \to \mathcal{C}$


1. creates any limits that $\mathcal{C}$ admits, without any assumption on $T$, and
2. creates colimits of any shape $X$ such that $\mathcal{C}$ admits $X$-shaped colimits and the underlying functor of $T$ preserves them.


These are Theorem 5.7 and Corollary 5.5 in Emily Riehl and Dominic Verity's paper [Completeness results for quasi-categories of algebras, homotopy limits, and related general constructions](http://arxiv.org/abs/1401.6247).


If $\mathcal{C}$ is stable and the underlying functor of $T$ is exact, 1 and 2 tell you that a pushout square of algebras is a pushout square in $\mathcal{C}$ is a pullback square in $\mathcal{C}$ is a pullback square of algebras; so the $\infty$-category of algebras of $T$ is also stable.


EDIT: About triangulated categories I don't know, but take a look at Paul Balmer's [Separability and triangulated categories](http://www.math.ucla.edu/~balmer/research/Pubfile/Sep-Triang.pdf), which seems to contain a similar result for something called separable monads.


