Density of set of all sequences with partial sum $\Theta(n)$ in space $\ell_\infty$ Let $\mathcal S$ be the subset of $\ell_\infty$ consisting of sequences $z$ with the following property
$z\in \mathcal S$ iff $\exists C\in\mathbb{R}$ such that the sequence $y$ given by
$$y(n) = \sum_{i=1}^n z(i) - Cn$$ is bounded.
So $(1,1,1,1,\ldots)\in \mathcal S$, $(1,0,1,0,\ldots)\in \mathcal S$, but $(1,1/2,1/3,1/4,\ldots)\not\in\mathcal S$ because partial sums are growing as $\log n$.
Is there a way to decide whether $\mathcal S$ is dense in $\ell_\infty$?

For clarity, $\ell_\infty$ is set of functions $\mathbb{N}\to\mathbb{R}$, which are bounded in a usual sense, with $\sup$ norm.
 A: Consider a $z\in \ell^\infty$ that is $-1$ on the first entry, $1$ on the next nine entries, $-1$ on the next $90$ entries, $1$ on the next $900$ entries etc. Formally we put $z(i)=(-1)^k$ where $k$ is such that $10^{k}<i\leq 10^{k+1}$.
Now, for $n\geq 1$, put $k_n=10^n-1$. Clearly, for odd $n$, $$\sum\limits_{i=1}^{k_n}z(i)\geq \frac{k_n}{2}.$$
Whereas for even $n$,
$$\sum\limits_{i=1}^{k_n}z(i)\leq -\frac{k_n}{2}.$$
We claim that the ball $B(z; 1/4)$ does not intersect $\mathcal{S}$ which will show that $\mathcal{S}$ isn't dense in $\ell^\infty$.  So suppose $\tilde{z}\in \ell^\infty$ satisifes $\|\tilde{z}-z\|_\infty\leq1/4$ and fix a $C\in\mathbb{R}$. For odd $n$ we have,
$$\sum\limits_{i=1}^{k_n}\tilde{z}(i)-Ck_n\geq \sum\limits_{i=1}^{k_n}z(i)-\frac{k_n}{4}-Ck_n\geq \frac{k_n}{2}-\frac{k_n}{4}-Ck_n=k_n(\frac{1}{4}-C).$$
In particular, the only way $\sum\limits_{i=1}^{k_n}\tilde{z}(i)-Ck_n$ can be bounded for odd $n$ is if $C\geq 1/4$. On the other hand, for even $n$,
$$\sum\limits_{i=1}^{k_n}\tilde{z}(i)-Ck_n\leq \sum\limits_{i=1}^{k_n}z(i)+\frac{k_n}{4}-Ck_n\leq -\frac{k_n}{2}+\frac{k_n}{4}-Ck_n=k_n(-\frac{1}{4}-C).$$
In particular the only way $\sum\limits_{i=1}^{k_n}\tilde{z}(i)-Ck_n$ can be bounded for even $n$ is if $C\leq -1/4$. Thus there is no choice of $C$ will make the values $\sum\limits_{i=1}^{n}\tilde{z}(i)-Cn$ uniformly bounded in $n$.
UPDATE: A much simpler argument would be that $\mathcal{S}$ is a proper subspace of $\ell^\infty$ and therefore is nowhere dense. In particular, it isn’t dense.
