A formula for the isomorphism $\ell^1\to c^*$

Question

For every bdd sequence $t=(t_n)$, let's put $\phi_t:c\to \mathbb{C}$ given by $(x_n)\to \sum x_nt_n$. Clearly $t\to \phi_t$ serves a formula for the isometrically isomorphism $\ell^1\simeq c_0^*$. What about the formual of $\ell^1\simeq c^*$?

Answer 1

A concise way is

$$(a_n)_{n=0}^\infty \in \ell^1 \mapsto f \in c^* $$where$$ f\left((x_n)_{n=1}^\infty\right) = a_0 \cdot \left(\lim_{n\to\infty} x_n\right) + \sum_{n=1}^\infty a_nx_n, \quad\forall (x_n)_{n=1}^\infty \in c$$

Answer 2

For sequences starting with $1$, you can use $$ (x_n) \mapsto \sum_{n = 1}^n x_n t_{n+1} + t_1 \, \lim_{n\to\infty} x_n.$$