James space is Banach. I have a problem with the proof of the completeness of James space $(J,\|\cdot\|_J)$.
$ J = \{ x \in c_0 : \|x\|_J < \infty \} $, where $ c_0 = \{ x = (x^1, x^2, ...) : \lim_{n \to \infty}{x^n} = 0\}$
$ \|x\|_J := \sup_{ n_1 < ...<n_k }( \Sigma_{i=1}^{k-1} (x_{n_i} - x_{n_i+1})^{2} )^{1/2} $
I supposed that $ x_j \in J $ is Cauchy, then $ \forall \epsilon>0 \exists N>0 : \forall j,p \geq N  \|x_j - x_p\|_J < \epsilon$
And then I consider $ |x_j^{k}| = \lim_{n \to \infty}|x_j^{k} - x_j^{n}| $, but I stuck with that.
Anyone have some ideas how to prove that $(J,\|\cdot\|_J)$ is a Banach?
 A: An equivalent condition for completeness is that every norm convergent series is convergent. i.e. if for some $x_{n}$'s in a space $X$, you have $\sum_{n=1}^{\infty}||x_{n}||<\infty$ then $\lim_{n\to\infty}\sum_{k=1}^{n}x_{n}$ must belong to $X$.
So assume that you have $\sum_{n=1}^{\infty}||x_{n}||_{J}<\infty$. Then you have $|x_{n}^{k} -x_{n}^{m}|\leq ||x_{n}||_{J}$ for the choice of $n_{1}=n$ , $n_{2}=m$ .
This holds for all $m\geq k+1$.
Thus letting $m\to\infty$ we have $|x_{n}^{k}|\leq ||x_{n}||_{J}$ and hence $\sum_{n=1}^{\infty}|x_{n}^{k}|<\infty$ .
Define $y^{k}=\sum_{n=1}^{\infty}x_{n}^{k}$ for all $k\in\Bbb{N}$ .
Then $\sum_{n=1}^{\infty} x_{n}=(\sum_{k=1}^{\infty}x^{1}_{n},\sum_{k=1}^{\infty}x^{2}_{n},...)=(y^{1},y^{2},....)$
So it only remains to show that $y=(y^{1},y^{2},...)\in J$ .
Fix $n_{1},...n_{k}$.
Then $$\bigg(\sum_{i=1}^{k}(y^{n_{i}}-y^{n_{i+1}})^{2}\bigg)^{\frac{1}{2}}=\bigg(\sum_{i=1}^{k}\bigg(\sum_{n=1}^{\infty}\bigg(x_{n}^{n_{i}}-x_{n}^{n_{i+1}}\bigg)\bigg)^{2}\bigg)^{\frac{1}{2}}\\\leq \sum_{n=1}^{\infty}\bigg(\sum_{i=1}^{k}(x_{n}^{n_{i}}-x_{n}^{n_{i+1}})^{2}\bigg)^{\frac{1}{2}}\leq \sum_{n=1}^{\infty}||x_{n}||_{J}$$ .
This holds for all such $n_{i}$'s and $k$.
And hence $||y||_{J}\leq \sum_{n=1}^{\infty}||x_{n}||_{J}<\infty$
Hence $y\in J$.
Thus $J$ is a Banach space.
