Convergence and bounds for $\sum_{k=1}^{n}\left(\frac{\phi\left(k\right)}{k}\log\left(\frac{2^{k}}{2^{k}-1}\right)\right)$

Given that $\phi(k)$ denotes Euler's totient function for $k$, we have this inequality

$$1-\frac{1}{2^n}<\sum_{k=1}^{n}\left(\frac{\phi\left(k\right)}{k}\log\left(\frac{2^{k}}{2^{k}-1}\right)\right)<1$$

for any positive integer $n$ (and $\log$ denotes the natural log). How would I go about proving this? I found this problem while digging around on my summer camp's homework sets that I unfortunately wasn't able to finish off.

I know that for $n = 1$, \begin{align*} \sum_{k=1}^{1}\left(\frac{\phi\left(k\right)}{k}\log\left(\frac{2^{k}}{2^{k}-1}\right)\right) &= \frac{\phi(1)}{1}\log\left(\frac{2^1}{2^1 - 1}\right) \\ &= \log(2) \end{align*} which is greater than $\frac{1}{2}$. How would I go about expanding and simplifying further terms, though? (and would induction work for this problem?)

$$\sum_{d | n} \phi(d)=\sum_{d| n}\sum_{m\le n, \gcd(m,n)=\frac{n}d} 1 = n$$ and

$$\sum_{k\le K}\frac{\phi(k)}{k}\log(\frac{2^k}{2^k-1}) = \sum_{k\le K} \frac{\phi(k)}{k}\sum_{m\ge 1} \frac{2^{-km}}{m}$$ $$ =\sum_{n\ge 1} 2^{-n} \sum_{d| n, d\le K} \frac{\phi(d)}{n}\ \ \begin{array} \ \ge \sum_{n=1}^K 2^{-n}\\ \le \sum_{n=1}^\infty 2^{-n}\end{array} $$