Meager set of number in base $2$ For each real $x \in (0,1]$ let $(d_n(x))_{n\ge 1}$ be its unique binary expansion with infinitely many $1$, i.e.,
$$
x=\sum_{n\ge 1}\frac{d_n(x)}{2^n}.
$$

Question. Is it true that the set
  $$
\left\{x \in (0,1]: s(x)=\sum_{n:\, d_{2n}(x)=1}\frac{1}{n} \le 1\right\}
$$
  is meager?

 A: I assume you are endowing the space $X=(0,1]$ with the usual topology from the real line. Then the answer is Yes, your set (call it $M$ ) is meager. 
This is because it is a nowhere dense set, because it is closed and has an empty interior. 
Consider any point $x\in M$.  Consider a ball $B$ of radius $\frac{1}{2^{2N}}$ around $x$. Then $B$ includes all points for which the first $2N$ binary digits match ṭhose of $x$.  In particular, it includes $y$ whose digits $d_{2m}$ are all 1 for $N<m\leq P$ where $P$ is such that the sum $\sum_{m=N+1}^P \frac{1}{m} > 1$. That is, $y\notin M$. So $M$ contains no open balls and so its interior is empty. 
Now consider $ x \in U = X\setminus M$. Since $s(x)>1$, let $g=\frac{1+s(x)}{2}$. Then $g>1$ and  we can find some natural number $N$ such that $\sum_{n:n\leq N, d_{2n}=1} \frac{1}{n}>g$. Consider the set of all $y$ whose digits match those of $x$ upto  $N$ digits. This set is the ball $B$ of radius $\frac{1}{2^{2N}}$ centered at $x$. It's clear that $\forall y\in B, s(y)>g$ since the first $2N$ digits of $y$ are same as those of $x$. That is, $B \subset U$. Since we can find such an open ball contained within $U$ for all $x\in U$, we know that $U$ is open. In other words, $M$ is closed. 
