In a Banach space, absolute convergence of series implies convergence

Question

Proof/Hint Request :

I came upon a shortly stated Lemma while revising for my Functional Analysis semester exam. It follows as :

Lemma : The absolute convergence of $\sum_{n=1}^{+ \infty}x_n$ implies the converge of $\sum_{n=1}^{+\infty} x_n$, where $x_n$ is a sequence in a Banach Space.

Now, I know that the absolute convergence of the noted sequence means that $\sum_{n=1}^{+ \infty} \|x_n\|$ is convergent, but I cannot see how to find a way to prove the lemma.

Any hints or elaborations will be appreciated.

Answer 1

Hint: Use the triangle inequality: $\mid S_n-S_m\mid=\mid x_n+x_{n-1}+\dots +x_{m+1}\mid\le \mid x_n\mid+\dots+\mid x_{m+1}\mid$.

Then $S_n$ will be Cauchy, hence convergent.

Answer 2

I believe this is incorrect in general - you need to be working over a Banach space. As for a hint when this is true:

Suppose $(x_n)$ is absolutely summable and consider the difference $$ |a_n-a_m|,$$ where $a_k = \sum_{j=0}^k x_j$. Using absolute summability can you show that $a_n$ is Cauchy?

Edit: thinking about it, the property of absolute convergence implying convergence (for sums) should be equivalent to the Banach property. As a challenge maybe try to prove the opposite implication than the one in this question.