I would like to know if the following is true : consider the series $\sum_{n\geq 1}x_n$ in a Banach space, denote $s_n= \sum_{k=1}^{n}x_k$ the partial sum and assume that $\sum_{n\geq 1}\lVert x_n\rVert < \infty$. Then the series $\sum_{n\geq 1}x_n$ is convergent and
$ \left\lVert\sum_{n\geq 1}x_n\right\rVert\leq\sum_{n\geq 1}\lVert x_n\rVert $
I think this is true because : the sequence $v_n =\sum_{k=1}^{n}\lVert x_k\rVert $ converge in $\mathbb{R}$ since it is an increasing and bounded sequence and therefore it is a Cauchy sequence. Now consider $\epsilon>0$, there exists $N\in\mathbb{N}$ such that
$ m>n>N\implies \lVert v_m - v_n \rVert <\epsilon $
Since $\lVert s_m - s_n\rVert = \lVert \sum_{k=n+1}^{m}x_k\rVert \leq\sum_{k=n+1}^{m}\lVert x_k\rVert < \epsilon$ this shows that $s_n$ is also a Cauchy sequence, since we are in a Banach space, this shows the convergence of $s_n$ and therefore of the series $\sum_{n\geq 1}x_n$. The last inequality just follow from the triangle inequality and by taking the limit.
Is this seems right ? Don't hesitate to propose your proof !
Thank you !
That looks perfect to me. In fact the reciprocal is true, a normed space is Banach if and only if every convergent sequence is absolutely convergent. Take a Cauchy sequence $(x_k)\subset E$. Then you can find a subsequence $(x_{n_k})$ such that $||x_{n_{k+1}}-x_{n_k}|| < 2^{-k}$. Let $y_k = x_{n_{k+1}}-x_{n_k}$. Then $\sum ||y_k||<\sum 2^{-k}<+\infty$, is absolutely convergent and, by hypothesys, it converges. But $\sum_{k=1}^N y_k = x_{n_N} - x_{n_1}$ and hence $x_k$ has a convergent subsequence. But since $x_k$ is Cauchy, the whole sequence is convergent.