Sequence of partial sums converge Let $(X,\|.\|)$ be a Banach space and a sequence $(x_n) \subseteq X$ be such that the series $\sum_{n=1}^{\infty} \|x_n\|$ converges in $\mathbb R$. Prove that the sequence $S_n=x_1+...+x_n$ converges in $(X,\|.\|)$.
This seems very obvious but I can't explain. It is said $\sum_{n=1}^{\infty}\|x_m\|$ converges so that means every term in the $S_n$ converges so the sum of these convergence limits add up to a finite number so $S_n$ converges.
This is what I think.
But I don't know how to actually answer it.
 A: Hint : try to prove that $S_n$ is a Cauchy sequence. As $(X, \| \cdot \|)$ is a Banach space, you will be able to conclude that $S_n$ converges in $X$.

 Let $\epsilon >0$. We want to find $N>0$ such that for all $n>m>N$, $\|S_n-S_m\|<\epsilon$. But $\|S_n-S_m\| = \|x_{m+1}+\cdots + x_n\|$... so you can use the triangle inequality, and your hypothesis, which says that there is an integer $N'>0$ such that $\sum_{n≥N'} \|x_n\|<\epsilon$.

More hints:

 You can take $N'=N$. Indeed, if $n>m>N$ then $$\|S_n-S_m\| = \|x_{m+1}+\cdots + x_n\| \leq \sum_{k=m+1}^n \|x_k\| \leq \sum_{k≥m+1} \|x_k\| \leq \sum_{k≥N} \|x_k\| < \epsilon$$

NB : your have proved that the absolute convergence implies the convergence, in every Banach space. Actually, it is possible to show that if a normed space satisfies the property "absolute convergence $\implies$ convergence" , then it is a Banach space!

To answer your comment more precisely:
This is because $A_K = \sum_{n=1}^{K}\|x_n\|$ converges to $L =\sum_{n=1}^{\infty}\|x_n\|$ when $K \to \infty$.
Then, there exists $N'=K>0$ such that 
$$|A_{k-1}-L|=\left|\sum_{n=k}^{\infty}\|x_n\| \right|<\epsilon$$ 
for all $k≥N'=K$. In particular this holds for $k=N'$, so that 
$$\left|\sum_{n=N'}^{\infty}\|x_n\| \right|<\epsilon$$
as desired.
