What is wrong with this "counterexample" of boundedness of weakly convergent sequences?

Question

Weakly convergence sequences $\{u_n\}$ in a Hilbert space $H$ are bounded.

Here is an attempted "counterexample". What is wrong with this?

Let $H = \ell_2(\mathbb{N})$, and let $\{e_n\}$ be the standard basis. Let a sequence $\{u_n\}\subset \ell_2$ be given by $$ u_n = n^{1/4} e_n. $$ This sequence is not bounded, since $\|u_n\| = n^{1/4}$. But $u_n\to 0$ weakly, as seen as follows: let $v\in \ell_2$ be arbitrary, $v = \sum_k v_k e_k$. Since $\sum_k|v_k|^2<+\infty$, there exists a $C>0$ such that for all $k$, $v_k< Ck^{-1/2}$. Now, $$ \langle v,u_n\rangle = v_n n^{1/4} < C n^{-1/4} \to 0. $$ Thus, $u_n\to 0$ weakly in $\ell_2$.

Answer 1

The mistake is in the conclusion $|v_k|<Ck^{-1/2}$. A sequence in $v\in \ell_2$ must tend to $0$, but it can do so arbitrarily slowly. I'll give another form of martini's example: $$ v_k = \begin{cases} 1/n,\quad &\text{ if $k=2^n$ for some $n\in \mathbb N$} \\ 0 ,\quad &\text{ otherwise} \end{cases} $$ This sequence is in $\ell_2$ but for every $\epsilon>0$, $$\limsup_{k\to\infty} k^\epsilon |v_k|=\infty$$