Why is the room of all weak p summable $(x(k))_k$ is weak p summable means $\forall x' \in X': \sum_k |{x'(x(k))}|^p < \infty$
$l_p^w(X)$={all weak p summable sequences in X}
$||x||_{l_p^w}=\underbrace{\sup}_{||x'||=1} (\sum_{k=1}^{\infty} |x'(x(k)|^p)^\frac{1}{p}$.
My question is why is $l_p^w$ complete?

Some hints: consider $\left(x^{(n)}\right)_{n\geqslant 1}$ a Cauchy sequence in $\ell^w_p$. Using the inequality 
$$
\sup_{\left\lVert x'\right\rVert_{X'}=1}\left\lvert x'\left(x\left(k\right)\right)\right\rvert\leqslant\left\lVert x\right\rVert_{\ell^w_p}
$$
(valid for any fixed $k$) and applied to $x=x^{(n)}-x^{(m)}$, and using 
$$
\left\lVert x\right\rVert=\sup_{\left\lVert x'\right\rVert_{X'}=1}\left\lvert x'\left(x \right)\right\rvert,
$$
we derive that the sequence $\left(x^{(n)}\left(k\right)\right)_{n\geqslant 1}$ is Cauchy in $X$ for any $k$. We have to assume that $X$ is complete. Denote $x(k)$ the limit of $\left(x^{(n)}\left(k\right)\right)_{n\geqslant 1}$. We have to show that the sequence $\left(x\left(k\right)\right)_{k\geqslant 1}$ is in $\ell^w_p$ and is the limit of $\left(x^{(n)}\right)_{n\geqslant 1}$ in $\ell^w_p$. To this aim, fix a positive $\varepsilon$. There is $N$ such that if $m,n\geqslant N$, then $$\left\lVert x^{(m)}-x^{(n)}\right\rVert_{\ell^w_p}\leqslant \varepsilon.$$
In particular, for any integer $K$ and and $x'\in X'$ of norm $1$, 
$$
\sum_{k=1}^K\left\lvert x'\left(x^{(m)}\left(k\right)\right)-x'\left(x^{(n)}\left(k\right)\right)\right\rvert^p\leqslant \varepsilon^p.
$$
Let $n$ going to infinity, then $K\to +\infty$ and after the supremum over $x'$. 
