weak implies strong convergence in $l^1$

Question

I want to prove Schur theorem stating that every weak convergent sequence in $l^1$ is strongly convergent. If we let $(x_n)_n=(x^i_n)\in l^1$, then $(x_n)_n$ converge weakly to $x$ if for any $y \in l^\infty$ we have $$(x_n-x,y)_{l^1\times l^\infty}\rightarrow 0.$$ Now, let $y=sing(x_n-x)$, we obtain $$(x_n-x,x_n-x)_{l^1\times l^\infty}=||x_n-x||_{l^1}\rightarrow 0.$$ I this passage is correct?. Thank you.