Show that $(y_n)\in l^q$

Let $x=(x_n)$ and define $\phi_j:l^p\to\mathbb{R}$ by $$\phi_j(x)=\sum_{n=1}^{j}x_ny_n.$$ Note that $\phi_j\in (l^p)'$ for all $j\in\mathbb{N}$.

By hypothesis, is well define the functional $\phi:l^p\to\mathbb{R}$ by $$\phi(x)=\sum_{n=1}^{\infty}x_ny_n\hspace{2cm}(1)$$

As $\lim_{j\to\infty}\phi_j(x)=\phi(x)$ for all $x\in l^p$, using the Uniform Boundedness Principle, we can conclude that $$\phi\in (l^p)'$$ Now, as $(l^p)'=l^q$, we know that there's a $(z_n)\in l^q$ such that $$\phi(x)=\sum_{n=1}^{\infty}x_n z_n\hspace{2cm}(2)$$

By $(1)$ and $(2)$, we conclude that $(y_n)=(z_n)\in l^q$.