How to show that the sequence $(x^{(n)})$ weakly convergent in $l_p$, $1\le p\lt \infty$

Question

How to show that the sequence $(x^{(n)})$ weakly convergent in $l_p$, $1\le p\lt \infty$.

where $(x^{(n)})=(\underbrace{0,0,..0}_{n-1},1/n,1/(n+1),...,1/(2n),0,0,...)$ for $n\in\mathbb{N}$

Answer 1

Define for an integer $k$ the linear functional $L_k\colon x\mapsto $ the $k$-element of the sequence $(x_j)_{j\geqslant 1}$. Since $\lim_{n\to\infty}L_k(x^{(n)})=0$ for each $k$, if $(x^{(n)})_{n\geqslant 1}$ is weakly convergent, the limit is the null sequence.

Using the linear bounded function $L\colon\ell^1\to\mathbb R$ given by $L(x)=\sum_{j=1}^{+\infty}x_j$, one can see that $L(x^{(n)})\geqslant 1/2$ for each $n$, hence $(x^{(n)})_{n\geqslant 1}$ is not weakly convergent in $\ell^1$.

Since for $p>1$ the series $\sum_{n\geqslant 1}n^{-p}$ is convergent, we obtain that $\lVert x^{(n)}\rVert_p\to 0$, hence the sequence $(x^{(n)})_{n\geqslant 1}$ is norm convergent in $\ell^p$.