Not a basis for $ l^\infty$ then what is it? We know that $ l^\infty$ has not a Schauder basis and its Hamel basis is uncountably infinite.
Let $e_n=(e_{n1}, e_{n2},...)$ (for each $n\in \mathbb{N}$) s.t. $e_{nj}=0$ when $n\neq j$ and $e_{nn}=1$, now for $x=(x_1,x_2,\ldots)\in l^\infty$ we have:
$$x=\sum_{n=1}^\infty x_ne_n$$
$\{ e_n \}_{n\in\mathbb{N}}$ is not a Hamel basis and a Schauder basis then what is it?
 A: Note that "$x = \sum_n x_n e_n$" does not hold in norm if $x \in \ell^\infty \setminus c_0$, neither does it weakly, as for $x \not\in c_0$ there is some $x^*\in (\ell^\infty)^*$ with $x^*|_{c_0} = 0$ and $x^*(x) = 1$. Then $$ x^*\left(\sum_{n=1}^N x_ne_n\right) = 0 \not\to 1 = x^*(x) $$
It holds only weakly$^*$, as given $y \in \ell^1$, we have $xy\in \ell^1$ and hence
$$ \sum_{n=1}^N x_ny_n \to \sum_{n=1}^\infty x_n y_n $$
So $(e_n)$ is a basis only in weakly$^*$ sense.
A: It is not true that $$x=\sum_{n=1}^{\infty} x_ne_n$$ for $x\in\ell^{\infty}$, because convergence does not necessarily occur in the $\ell^{\infty}$-norm. Indeed, if  $x=(1,1,1,\ldots)$, then, for any $N\in\mathbb N$, $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(\underbrace{0,\ldots,0}_{N\text{ times}},1,1,1,\ldots)\|_{\infty}=1,$$ so $x$ cannot be the norm limit of the partial sums $\{\sum_{n=1}^{N}x_ne_n\}_{N\in\mathbb N}$. Therefore,  $\{e_n\}_{n\in\mathbb N}$ is not a Schauder basis.
But even if it were a Schauder basis, it would still not be a Hamel basis, because this would require that any element $x\in\ell^{\infty}$ must be a finite linear combination of elements of $\{e_n\}_{n\in\mathbb N}$, not merely a norm limit of the partial sums.
A: The sum you give is not norm convergent, unless $x\in c_0$. (It is weakly$\!^*$ convergent, but that is beside the point.) You have here a Schauder basis for $c_0$, a (comparatively) tiny subspace of $l^\infty$.
