Let $X$ be a normed space. I want to show that $X$ is a Banach space if and only if for all sequences $\{x_n\}\in X$ such that $\sum_n||x_n|| < \infty$ there exists an element $y\in X$ such that $$ \lim_{n\to \infty} ||y - (x_1+\dots+x_n)||=0. $$
- $=>$ First showing that $X$ is a Banach space. Let $\{x_n\}$ be a Cauchy sequence in $X$ such that $\sum_n ||x_n|| < \infty$ and choose $X \ni x = y - (x_1+ \dots + x_n)$ as a candidate limit for the Cauchy sequence $\{x_n\}$. Then we have $$ \begin{align} \lim_{n\to \infty}||x - x_n|| & = \lim_{n\to \infty}||y - (x_1+ \dots + x_n) - x_n|| \\ & \le \lim_{n\to \infty}||y - (x_1+ \dots + x_n)|| + \lim_{n\to \infty}||x_n|| \\ & \le 0 + 0 = 0, \end{align} $$ where $\lim_{n\to \infty}||x_n|| = 0$ as $\sum_n ||x_n|| < \infty$. But it seems I missed something here as to show $X$ is complete we need to be able to find a limit $x\in X$ for all Cauchy sequences, not just ones satisfying $\sum_n ||x_n|| < \infty$. So how can I fix this proof?
- $<=$ Now to show the other direction. Let $X$ be a Banach space and let $\{x_n\}$ be a sequence in $X$ such that $\sum_n||x_n|| < \infty$. This means that $\lim_{n\to\infty}||x_n|| = 0$. Set $y = x_1 + \dots + x_{n-1}$. Then we have $$ \begin{align} \lim_{n\to\infty}||y-(x_1+\dots+x_n)|| & = \lim_{n\to\infty}||(x_1 + \dots + x_{n-1}) - (x_1 + \dots + x_n)|| \\ & = \lim_{n\to\infty}||x_n|| = 0. \end{align} $$ I didn't use the completeness of the Banach space here so I'm not sure if I missed something. Does this look correct?