TITLE: How can we prove $1>\sum_{k=1}^{n}\frac{\varphi{(k)}}{k}\ln{\frac{2^k}{2^k-1}}>1-\frac{1}{2^n}$
QUESTION [5 upvotes]: Let $\varphi$ be Euler's totient function, where $\varphi{(1)}=1$. Prove that for any postive integer $n$ we have
  $$1>\sum_{k=1}^{n}\left(\dfrac{\varphi{(k)}}{k}\ln{\dfrac{2^k}{2^k-1}}\right)>1-\dfrac{1}{2^n}$$

Maybe this problem can be solved using integrals.
$$\ln{\dfrac{2^k}{2^k-1}}=\ln{2^k}-\ln{(2^k-1)}$$
But then I don't know how to deal with this term:
$$\dfrac{\varphi{(k)}}{k}$$

REPLY [6 votes]: For the lower bound:
$$\begin{eqnarray*}\sum_{k=1}^{n}\frac{\phi(k)}{k}\log\left(\frac{2^k}{2^k-1}\right)&=&\sum_{k=1}^{n}\frac{\phi(k)}{k}\left(-\log\left(1-\frac{1}{2^k}\right)\right)\\
&=&\sum_{k=1}^{n}\sum_{r=1}^{+\infty}\frac{\phi(k)}{kr\cdot 2^{kr}}\\
&=&\sum_{s=1}^{+\infty}\frac{1}{s 2^s}\sum_{\substack{m\mid s\\m\leq n}}\phi(m)\\
&>&\sum_{s=1}^{n}\frac{1}{s 2^s}\sum_{m\mid s}\phi(m)\\
&=&\sum_{s=1}^{n}\frac{1}{2^s}\\
&=&1-\frac{1}{2^n}.
\end{eqnarray*}$$
The upper bound simply follows from
$$\sum_{\substack{m\mid s\\m\leq n}}\phi(m)\leq s.$$