How can we prove $\sum\limits_{n=1}^{k}\frac{f(m,n)}{n}\approx\frac{\varphi(m)}{m}(\ln(k\prod\limits_{p|m}p^{\frac{1}{p-1}})+\gamma)$? First we need to create a function
$$f(m,n)=\begin{cases}
1&\text{if $m,n$ - coprime}\\
0&\text{otherwise}\\
\end{cases}$$
Then we can be sure, that for large $k$ and $|\mu(m)|=1$
$$\sum\limits_{n=1}^{k}\frac{f(m,n)}{n}\approx\frac{\varphi(m)}{m}\left(\ln\left(k\prod\limits_{p|m}p^{\frac{1}{p-1}}\right)+\gamma\right)$$
How can we prove that?
If I made some mistakes, sorry for my English.
 A: $\sum_{n=1}^k \frac{1_{(m,n) = 1}}{n} = \sum_{n=1}^k \frac{1}{n}\sum_{d \mid (n,m)} \mu(d) = \sum_{n=1}^k \sum_{d \mid n, d \mid m} \mu(d) = \sum_{d \mid m} \mu(d)\sum_{n \le k, d \mid n} \frac{1}{n} = $
$\sum_{d \mid m} \mu(d)\sum_{t \le \frac{k}{d}} \frac{1}{dt} = \sum_{d \mid m} \frac{\mu(d)}{d} [ \log \frac{k}{d} + \gamma + O(\frac{d}{k})]$. 
Using $\sum_{d \mid m} \frac{\mu(d)}{d} = \frac{\phi(m)}{m}$, our sum is $\frac{\phi(m)}{m}\log k + \frac{\phi(m)}{m}\gamma + O(\frac{\tau(m)}{k}) - \sum_{d \mid m} \frac{\mu(d)\log d}{d}$. So we just need to estimate this last term. If $m = p_1\dots p_l$ is squarefree, then this sum is $\frac{\log p_1}{p_1}+\dots+\frac{\log p_l}{p_l} - \frac{\log(p_1p_2)}{p_1p_2}-\dots-\frac{\log(p_{l-1}p_l)}{p_{l-1}p_l} + \dots$. For each $i$, pairing up the $\log p_i$ terms gives $\frac{\log p_i}{p_i} \sum_{d \mid \frac{m}{p_i}} \frac{\mu(d)}{d} = \frac{\phi(m/p_i)}{m/p_i}\frac{\log p_i}{p_i} = \phi(m/p_i)\frac{\log p_i}{m} = \frac{\phi(m)}{p_i-1}\frac{\log p_i}{m} = \frac{\phi(m)}{m}\frac{\log p_i}{p_i-1}$. Summing over $i$ finishes the job. I'll let you think about the (similar) case for when $m$ is not square free.
