Bounding $\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$ I'm looking for a bound depending on $N$ of $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$.
The following holds $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}} = \sum_{k=1}^N \sum_{i=0}^\infty \frac{1}{2^{ki}}=\sum_{n=0}^\infty \sum_{\substack{q\geq 0 \\ 1\leq p \leq N\\pq=n}}\frac{1}{2^{pq}}\leq 2N$ which does not seem accurate.
Numerical trials suggest that $\displaystyle N-\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$ is a convergent sequence (it's easy to prove it is decreasing).

$$\sum_{k=1}^{N}\frac{1}{1-\frac{1}{2^k}}=\sum_{k=1}^{N}\left(1+\sum_{j\geq 1}\frac{1}{2^{kj}}\right)\leq N+1+\frac{1}{3}+\ldots+\frac{1}{2^N-1}\leq N+\frac{5}{3}.$$
