# Normed Space $X$ is complete iff every absolutely convergent series in $X$ converges

I'm studying functional analysis. I have trouble with the following proposition and its proof. Wonder if someone could help me with the following questions:

Proposition: A normed space $X$ is complete if, and only if, any absolutely convergent series in $X$ converges.

Proof: Let $X$ be a Banach space, and suppose $\sum_{n=1}\| x_n\|$ converges. Let $y_n = \sum_{n=1}^{N}\| x_n\|$ so that $M > N$

$$\| y_m - y_n\| = \|\sum_{n=N + 1}^{M} {x_n}\| \leq \sum_{n=N + 1}^{M} \|{x_n}\| \rightarrow 0 \> \> as N,M \rightarrow \infty$$

Hence, $y_n$ is Cauchy sequence in the complete space $X$, and so converges.

Question 1: A normed space $X$ is complete --> is that equivalent to saying --> Space $X$ is Banach space?

Question 2: $\sum_{n=1}\| x_n\|$ converges --> is that equivalent to saying --> $x_n$ is an absolutely convergent series that is convergent?

Question 2: The definition of "$x_n$ is an absolutely convergent series" is that $\sum_{n=1}^\infty \|x_n\|$ converges. This says nothing at all about whether $\sum_{n=1}^\infty x_n$ converges (i.e., whether it is a convergent series). The point of the proof is to show that assuming $X$ is complete, then $\sum_{n=1}^\infty x_n$ does converge under these circumstances.
(Incidentally, the proof appears to have a typo where it defines $y_n = \sum_{n=1}^{N}\| x_n\|$. That should instead be $y_N = \sum_{n=1}^{N} x_n$. Also, the proof does not prove the entire proposition, since it only proves the forward direction of the "if and only if".)