Proposition 8.8.1. Let $E$ be a metric TVS with its topology induced by the pseudonorm $\rho: E \to [0, \infty)$, then the following are equivalent:
$E$ is complete.
For any $\seq{x_n}\subset X$ with $\sum_{n \in \natp}\rho(x_{n}) < \infty$, $\limv{N}\sum_{n = 1}^{N} x_{n}$ exists in $E$.
Proof. $(2) \Rightarrow (1)$: Let $\seq{x_n}\subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k}\subset \natp$ such that for each $k \in \natp$, $\rho(x_{n_{k+1}}- x_{n_{k}}) < 2^{-k}$.
Let $x = x_{n_1}+ \limv{N}\sum_{k = 1}^{N} (x_{n_{k+1}}- x_{n_k})$, then $x = \limv{k}x_{n_k}\in E$. Since $\seq{x_n}$ is a Cauchy sequence that admits a convergent subsequence, it is convergent.$\square$