What is wrong with the following proof that the dual space of $l^\infty$ is $l^1$?
Consider the mapping $\phi:l^1\to (l^\infty)'$, given by $\phi(y):l^\infty\to\mathbb{C}$ and $\phi(y)(x)=\langle x,y\rangle$. This is evidently convergent for all $x$ and $y$ as $$|\langle x,y\rangle|=\left|\sum_j x_j\overline{y_j}\right|\leq\sum_j |x_j\overline{y_j}|=\sum_j |x_j||y_j|\leq\|x\|_\infty\|y\|_1.$$
It is also easy to show that $\phi$ is linear. Now to show that $\phi$ is injective consider $\phi(y)=0$, then $\phi(y)(x)=0$ for all $x$. Take $e_j=(0,...,0,1,0,...)$, then $0=\phi(y)(e_j)=y_j$, so $y_j=0$ for all $j$ and thus $y=0$.
I assume something goes wrong when showing surjectivity. Let $l:l^\infty\to \mathbb{C}$ be a bounded linear functional and let $l_j=l(e_j)$. Take $x\in l^\infty$ to be $$x_j=\frac{\overline{l_j}}{|l_j|}$$ if $l_j\neq 0$ and $x_j=0$ if $l_j=0$. $x$ evidently belongs to $l^\infty$ as $||x||_\infty\leq 1$. Further $$l(x)=\sum_j |l_j|$$ which must then converge and so $\tilde{l}=\{l_j\}_{j\in \mathbb{N}}$ must belong to $l^1$. Now $\phi(\tilde{l})=l$.