Show that the limit is zero

Question

I have to show that if $(x_n)$ is weakly convergent in $X$ then for any $a>1$ $$\lim_{n\rightarrow\infty}\frac{\|x_1+\dots + x_n\|}{n^a}=0$$

My attempt:

If $(x_n)$ is weakly convergent, then it is bounded by some constant $C>0$. So we have: $$\frac{\|x_1+\dots + x_n\|}{n^a}\le\frac{\|x_1\|+\dots + \|x_n\|}{n^a}\le \frac{nC}{n^a}=\frac{C}{n^{a-1}} \longrightarrow 0$$

Is this correct?

Answer 1

The attempt is correct and it actually extends to bounded sequences.

Notice that it wouldn't necessarily work with $a=1$ (for example with $X=\ell^2$ and $(x_n)_{n\geqslant 1}$ an orthonormal sequence).