On weak closure again. I worked in this problem too: 
On the weak closure
But after that I could not think of anything that can help me about the $l_p$ case. I mean, $\{ n^{1/p} e_n \}$ has $0$ as a weak accumulation point but no subsequence of this set weakly converges to $0$ (with $p \in [1,\infty)$) ? 
I tried to use something about my $l_2$ proof but it uses strongly facts about Hilbert spaces. 
For example, to prove that $0$ is an accumulation point. Given a weak neighborhood $W$ of  $0$ and a natural $n_0$ if I suppose that, for all $n \geq n_0$, is true that $\sqrt{n}e_n \notin W$ I could find an absurd using the definition of $W$. But I used the fact that $\{ e_n \}$ is a hilbert basis. 
So, do you have any hint about the $l_p$ case? 
obs: I'm still studying english, sorry about my errors =p
Thanks! 
 A: I believe it's more natural to  consider the sequence $\{ n^{1/q}e_n\}$, where ${1\over p}+{1\over q}=1$.  
In spirit, showing that $0$ is in the weak closure of this set is exactly the same  as in the case for a Hilbert space:
Recall that the dual of $\ell_p$ is $\ell_q$; and given $y=(y_1,y_2,\ldots)$ in $\ell_q$, its action on $x=(x_1,x_2,\ldots)$ in $\ell_p$ is
$$
y(x)=\sum_{i=1}^\infty y_ix_i.
$$
A weak nhood of $0$ in $\ell_p$ has the form
$$
O=\{ x\in\ell_p\mid  |x_i^*(x)|<\epsilon, 1\le i\le n\}
$$
for some positive integer $n$,  $\epsilon>0$ and elements $x_1^*$, $x_2^*$, $\ldots\,$, $x_n^*$ in $\ell_q$.
Now suppose such an $O$ is given and that  no element $m^{1/q}e_m$ is in $O$. Then for each $m$, it would follow that 
$$
\sum_{i=1}^n m |x_i^*(e_m)|^q=
\sum_{i=1}^n |x_i^*(m^{1/q}e_m)|^q\ge \max_{1\le i\le n} |x_i^*(m^{1/q}e_m)|^q\ge \epsilon^q.
$$
From this, we would then have
$$
\sum_{m=1}^\infty\sum_{i=1}^n|x_i^*(e_m)|^q \ge\sum_{m=1}^\infty {\epsilon^q\over m}=\infty.
$$
But
$$
\sum_{m=1}^\infty\sum_{i=1}^n |x_i^*(e_m)|^q = 
\sum_{i=1}^n \sum_{m=1}^\infty|x_i^*(e_m)|^q =
\sum_{i=1}^n\Vert x_i^*\Vert_q^q<\infty;
$$
and thus the supposition that no element $m^{1/q}e_m$ is in $O$ cannot hold.
To show that no subsequence of  $\{ n^{1/q}e_n\}$ converges to $0$, use the fact that   weakly convergent sequences are norm bounded.
