Context
I'm reading an introductory book to Hilbert spaces and came up on a defination of complete normed spaces. The theorem states
Theorem 1.5.2.
A normed space is complete if and only if every absolutely convergent series converges
Question
I'm confused as to why the convergence of an absolutely convergent sequence is a requirement for complete normed spaces. The Absolute Convergence Theorem states that every absolutely convergent series must converge so why is that a requirement for normed spaces to be complete? isn't this also true for normed spaces that are not complete?
A normed space is complete iff every absolutely convergent series converges.
Note that if $(x_n)_{n=1}^\infty\subset X$ satisfies $\sum_{n=1}^\infty\|x_n\|<\infty$, then the partial sum sequence $(s_N)_{N=1}^\infty=(\sum_{n=1}^N x_n)_{N=1}^\infty$ is Cauchy because $$\lim_{\min\{M,N\}\to\infty}\|s_N-s_M\| = \lim_{\min\{M,N\}\to\infty}\bigl\|\sum_{n=1+\min\{M,N\}}^{\max\{M,N\}}x_n\bigr\|\leqslant \lim_{\min\{M,N\}\to\infty}\sum_{n=1+\min\{M,N\}}^{\max\{M,N\}}\|x_n\|=0.$$ If $X$ is complete, then $s_N$ converges.
For the converse, suppose every absolutely convergent series is convergent. Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence. Pick $N_1<N_2<\ldots$ such that $m,n\geqslant N_k$ implies $\|x_m-x_n\|<2^{-k}.$ Note that $\|x_{N_1}\|+\sum_{k=1}^\infty x_{N_{k+1}}-x_{N_k}\|<\infty$, the partial sums $$s_k:=x_{N_1}+(x_{N_2}-x_{N_1})+\ldots + (x_{N_k}-x_{N_{k-1}})=x_{N_k}$$ converges to some limit $x$. Then one can show that the whole sequence $(x_n)_{n=1}^\infty$ also converges to this $x$, since $$\lim_n\|x-x_n\| =\lim_n\lim_k \|x_{N_k}-x_n\|=0,$$ by the Cauchy condition.
As an example, take the space $P$ of all polynomials on $[0,1]$ with norm $$\|p\|=\max_{x\in [0,1]}|p(x)|.$$ Then $$\sum_{n=0}^N \frac{x^n}{n!}$$ is Cauchy but not convergent. It's Cauchy because it's actually uniformly converging to $e^x$, which is not a polynomial.