Proving that $c' \cong \ell^1$

Question

I'm trying to prove that the map $T:x \in \ell^1 \mapsto Tx \in c'$ given by $$(Tx)(y)=x_1\lim_{n\to\infty}y_n + \sum_{k=1}^\infty x_{k+1}y_k$$ is an isometric isomorphism between $\ell^1$ and $c'$ (the dual of the subspace of $\ell^\infty$ consisting of all convergent sequences). I have already shown that $T$ is a bounded linear isometry. To prove the surjectivity, I found that given $y' \in c'$ the sequence $$x=\left(y'(1)-\sum_{n=1}^\infty y'(e_n),y'(e_1),y'(e_2),y'(e_3),\ldots\right)$$ satisfies $Tx=y$. I'm denoting by $1$ the sequence $(1)_{n \in \mathbb{N}}$. It is remaining to prove that $x$ indeed belongs to $\ell^1$, i.e., that $\Vert x \Vert_1 < \infty$. I see that $$\Vert x \Vert_1 = \vert y'(1) - \sum_{n=1}^\infty y'(e_n)\vert + \sum_{n=1}^\infty \vert y'(e_n)\vert$$ but I can't see why this should be finite. I feel I'm missing something that should be clear. I also would like to justify why $\sum_{n=1}^\infty y'(e_n)$ would be convergent since I'm using this to define $x$.