Before I begin, an apology: it's been a while since I've done much analysis, so I might be misusing (or just missing) terminology.

I have a chain complex $(V_\bullet,\partial)$ of topological vector spaces (in particular, the differential is continuous). My vector spaces are *Cauchy-complete* in the following sense: If $x_n \in V_i$ is a sequence of vectors of fixed degree such that $\lim_{\min(m,n) \to \infty} (x_n - x_m) = 0$, then $\lim_{n\to \infty} x_n$ converges.

Actually, I have a particular sequence $x_n \in V_0$ that I care about. Unfortunately, $\lim_{n\to \infty} x_n$ does not converge. But it is "a Cauchy sequence up to homotopy." Specifically, for any two $m,n$, I have a specific vector $y_{m,n} \in V_1$ such that: $$(*) \quad\quad\quad \lim_{\min(m,n) \to \infty} (x_n - x_m - \partial(y_{m,n})) = 0 $$ Also, I should mention that each individual $x_n$ is not closed, so I can't talk about their classes in homology. But I do know that $\lim_{n\to \infty}(\partial x_n) = 0$. (Going in the other direction, none of the natural limits of the $y_{m,n}$ converge, but I can prove variations of equation $(*)$ for them too, and so on ad infinitum.)

**My question is whether $\lim_{n\to \infty} x_n$ exists in some homotopical sense,** and how to define it. Of course, I don't expect that there is a specific (closed) element $x_\infty = \lim x_n$. But I would expect that there is some natural set of such elements $x_\alpha$, for $\alpha$ ranging over some indexing set, along with specific homotopies $y_{\alpha,\beta}$ satisfying $\partial(y_{\alpha,\beta}) = x_\beta - x_\alpha$.

**If not, are there additional reasonable conditions that would assure such a limit?** I have pretty good control over my particular example. For instance, much stronger than the Cauchy-completeness, I know that if $\lim_{n\to \infty} (x_{n+1} - x_n) = 0$, then $\lim_{n\to \infty} x_n$ converges. The reason I know things like that is because the specific vector spaces $V_\bullet$ that I care about are essentially power-series algebras over $\mathbb Q$. So if there's some natural condition that's necessary for convergence, I can easily check it, and it's probably satisfied.