Bounding $\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$

Question

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).

Answer 1

We have $$\sum_{k\leq N}\frac{2^{k}}{2^{k}-1}=N+\sum_{k\leq N}\frac{1}{2^{k}-1} $$ and recalling that holds the identity $$\sum_{k=1}^{\infty}\frac{1}{1-a^{k}}=\frac{\psi_{\frac{1}{a}}\left(1\right)+\log\left(a-1\right)+\log\left(\frac{1}{a}\right)}{\log\left(a\right)} $$ where $\psi_{q}\left(z\right) $ is the $q $-polygamma function, we have $$\sum_{k\leq N}\frac{2^{k}}{2^{k}-1}=N+\sum_{k\leq N}\frac{1}{2^{k}-1}\leq N+\sum_{k\geq1}\frac{1}{2^{k}-1}=$$ $$=N+1-\frac{\psi_{\frac{1}{2}}\left(1\right)}{\log\left(2\right)}\approx N+1.606695. $$

Answer 2

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

Answer 3

Note that $$\frac{1}{1-\frac{1}{2^k}} = 1+ \frac{1}{2^k-1}$$ so $$N \le \sum_{k=1}^N\frac{1}{1-\frac{1}{2^k}} \le N+ \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le N+2$$ since $1\le \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le 2$