Series criterion for Banach spaces

Assume $V$ is Banach. If $\sum x_n$ is absolutely convergent, [L2] makes its partial sums Cauchy, and [L1] then makes those partial sums converge in $V$. So every absolutely convergent series converges.

Assume [A1]. Let $(y_m)$ be a Cauchy sequence in $V$. Choose inductively a strictly increasing sequence $(n_k)$ with $\mathrm{\parallel }{y}_m-y_n\mathrm{\parallel }<2^{-k}$ whenever $m,n\ge n_k$. Then $\mathrm{\parallel }{y}_{n_{k+1}}-y_{n_k}\mathrm{\parallel }<2^{-k}$ for every $k$.

The series ${\sum }_{k=0}^{\mathrm{\infty }}(y_{n_{k+1}}-y_{n_k})$ is absolutely convergent because $\sum 2^{-k}$ converges and each term has norm at most $2^{-k}$. By [A1] it therefore converges to some $z\in V$.

Its partial sums are $y_{n_m}-y_{n_0}$, so the subsequence $(y_{n_m})$ converges to $y_{n_0}+z$.

Given $\epsilon >0$, choose $K$ so that $\mathrm{\parallel }{y}_m-y_n\mathrm{\parallel }<\epsilon \mathrm{/}2$ for $m,n\ge K$, and choose $m$ with $n_m\ge K$ and $\mathrm{\parallel }{y}_{n_m}-(y_{n_0}+z)\mathrm{\parallel }<\epsilon \mathrm{/}2$. Then for every $n\ge K$, $\mathrm{\parallel }{y}_n-(y_{n_0}+z)\mathrm{\parallel }\le \mathrm{\parallel }{y}_n-y_{n_m}\mathrm{\parallel }+\mathrm{\parallel }{y}_{n_m}-(y_{n_0}+z)\mathrm{\parallel }<\epsilon$. So $(y_n)$ converges, and $V$ is complete.

Step 1.1 proves $(1)\Rightarrow (2)$ and steps 1.2 through 4.1 prove $(2)\Rightarrow (1)$, so the two conditions are equivalent.