Asymptotic behavior of $\sum_{r=0}^\infty r\left[ \left(1-\frac 1{2^{r+1}}\right)^m-\left(1-\frac 1{2^{r}}\right)^m\right]$ as $m\to+\infty$. 
What is the asymptotic behavior of the following series as $m\to+\infty$?
$$S_m=\sum_{r=0}^\infty r\left[ \left(1-\frac 1{2^{r+1}}\right)^m-\left(1-\frac 1{2^{r}}\right)^m\right],\qquad m\in\mathbb N_+.$$

This problem comes from a friend of mine majoring in engineering science, who did some numerical simulations and told me it is likely to approach $\log_2m$.
After expanding the $(1-a)^m$ using Binomial formula and calculating the series $\sum_{r=0}^\infty rx^r$, I convert the above series to the following finite summation:
$$S_m=\sum_{k=1}^m {m\choose k}\frac{(-1)^{k-1}}{2^k-1},\qquad m\in\mathbb N_+. $$
Is there any method to see the asymptotic behavior of the above summation as $m\to+\infty$? I not only want to know whether it approches $\log_2m$, but also want to know the convergence rate. So it will be better to write down its asymptotic expansion at $m\to+\infty$.
Any help will be appreciated!
 A: Indeed we have
$$ S_m = \log_2 (m) + \mathcal{O}(1) $$
as $m\to\infty$.

Define $x_r$ by
$$ x_r = (1 - 2^{-r})^m. $$
Then $r = f_m(x_r)$ for $f_m$ defined by $f_m(x) = -\log_2(1 - x^{1/m})$. Now by noting that $f_m$ is increasing on $[0, 1)$, we may bound $S_m$ by
$$ S_m
= \sum_{r=1}^{\infty} f_m(x_r) (x_{r+1} - x_r)
\leq \int_{0}^{1} f_m(x) \, \mathrm{d}x. $$
Similarly,
$$ S_m + 1
= \sum_{r=0}^{\infty} (r + 1) (x_{r+1} - x_r)
= \sum_{r=0}^{\infty} f_m(x_{r+1}) (x_{r+1} - x_r)
\geq \int_{0}^{1} f_m(x) \, \mathrm{d}x. $$
Finally, the integral is computed as
\begin{align*}
\int_{0}^{1} f_m(x) \, \mathrm{d}x
&= -\int_{0}^{1} mx^{m-1} \log_2(1-x) \, \mathrm{d}x \\
&= \left[ (1-x^m) \log_2(1-x) \right]_{0}^{1} + \frac{1}{\log 2}\int_{0}^{1} \frac{1-x^m}{1-x} \, \mathrm{d}x \\
&= \frac{H_m}{\log 2},
\end{align*}
where $H_m = 1 + \frac{1}{2} + \cdots + \frac{1}{m}$ is the harmonic number. Therefore, by using the bound $\log (m) \leq H_m \leq \log (m) + 1$, we conclude
$$ \log_2 (m) - 1 \leq S_m \leq \log_2 m + \frac{1}{\log 2}. $$
A: Inspired by @Sangchul Lee's elegant solution, a very good approximation is given by
$$S_m\sim\frac{H_m}{\log (2)}-\frac{1}{2}$$ For $m=100$, the difference between rhs and lhs is $2.28\times 10^{-6}$ and for $m=1000$, it is $1.01\times 10^{-6}$.
Using the sharps bounds given here
$$ \log
   (m)+\frac{1}{2 m}+\gamma-\frac{1}{12 m^2+\frac{2 (7-12 \gamma )}{2 \gamma -1}} \leq H_m$$
$$H_m \leq \log
   (m)+\frac{1}{2 m}+\gamma-\frac{1}{12 m^2+\frac{6}{5}}$$ we have a quite good approximation and bounds.
