Let $X$ be a normed space. Prove that $X$ is complete if and only if $\Sigma_{n=1}^{\infty} x_n$ exists for any sequence $\{x_n\}$ that satisfies $\Sigma_{n = 1}^\infty \|x_n\| < \infty$. Here $\Sigma_{n=1}^{\infty} x_n$ means the limit of $\Sigma_{n=1}^{N} x_n $ as $N \rightarrow \infty$
I have started to learn Introduction to functional analysis, but I am not sure if I understand correctly. Could you give me some answers (or hints) to the question? Thank you.
Recall the definition of completeness is that every Cauchy sequence has a limit in $X$. So for the forward direction, if you assume $X$ is complete you want to show that $\left\{\sum_{n=1}^Nx_n\right\}_{N=1}^\infty$ is a Cauchy sequence if the the sequence of norms $\left\{\sum_{n=1}^N\|x_n\|\right\}_{N=1}^\infty$ is convergent i.e. the series of norms $\sum_{n=1}^\infty \|x_n\|$ is convergent. (Hint: think about the Cauchy criterion for series convergence).
In the other direction, if you assume that any sequence whose series of norms converges has a limit in $X$, you must show that any Cauchy sequence in $X$ has a limit in $X$.
