Dual space of $l^1$

Hint.

Step I. $\ell^1 \subset \ell^\infty$. This is clear as every bounded sequence (i.e., member of $\ell^\infty$) defines a bounded linear functional on $\ell^1$.

Step II. If $\varphi\in(\ell^1)^*$, and $e_n=(0,0,\ldots,1,\ldots)\in\ell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set $$ u_n=\varphi(e_n). $$ Then $$ \lvert u_n\rvert=\lvert\varphi(e_n)\rvert \le \|\varphi\|_{(\ell^1)^*} \|e_n\|_{\ell^1}=\|\varphi\|_{(\ell^1)^*}, $$ and hence $\{u_n\}$ is a bounded sequence, i.e., $\{u_n\}\in\ell^\infty$.

Step III. It remains to show that $\varphi(x)=\sum_{n=1}^\infty u_nx_n$, for all $x=\{x_n\}\in\ell^1$.