Estimation of $\sum_{n \leq x} \frac{k(n)}{n}$ , with $k(n)$ the squarefree kernel I came across a poblem where they ask you to find an estimation of $\sum_{n \leq x} \frac{k(n)}{n}$, with $k(n) = \prod_{p \mid n} p$ the squarefree kernel of $n$, with an error term of $O(\sqrt{x})$.
I've worked on it for a while now, but I'm not able to find a solution. Does somebody know a solution?
 A: This is an exercise from Tenenbaum's 'Introduction to Analytic and Probablistic Number Theory'  It is #11 from page 54. It has a hint which is a preceding exercise. 
First consider a decomposition of power-full number($p|n\Rightarrow p^2|n$). 
If $n$ is power-full, then $n$ can be written uniquely as $n=m^3d^2$ where $m$ is square free. (This is from preceding exercise about power-full numbers)
Also, any natural number $n$ can be written uniquely as $n=m m_f$ where $m$ is square free, and $m_f$ is power-full. That forces $(m,m_f)=1$.  
Now, we start from decomposing the sum into sums over $m$ and $m_f$:
$$
\sum_{n\leq x}\frac{k(n)}{n}=\sum_{m_f\leq x}\sum_{\substack{{m\leq \frac{x}{m_f}} \\\  {(m,m_f)=1}}} \frac{k(mm_f)}{mm_f}$$
$$=\sum_{m_f\leq x}\sum_{\substack{{m\leq \frac{x}{m_f}} \\\  {(m,m_f)=1}}} \frac{k(m_f)}{m_f}$$
$$=\sum_{m_f\leq x}\frac{k(m_f)}{m_f}\sum_{\substack{{m\leq \frac{x}{m_f}} \\\  {(m,m_f)=1}}} 1$$
By the way, we have the following
$$
\sum_{\substack{{n\leq x} \\\ {n: \textrm{ square free} }\\\ {(n,q)=1} }}1 = \frac{6}{\pi^2}\prod_{p|q} \left(1+\frac{1}{p}\right)^{-1} x +O(\sqrt x)$$
We plug this in place of the sum over $m$: 
$$
\sum_{m_f\leq x}\frac{k(m_f)}{m_f}\sum_{\substack{{m\leq \frac{x}{m_f}} \\\  {(m,m_f)=1}}} 1=\sum_{m_f\leq x}\frac{k(m_f)}{m_f}\left(\frac{6}{\pi^2}\prod_{p|m_f}\left(1+\frac 1 p\right)^{-1}\frac{x}{m_f}+O(\frac{\sqrt x}{\sqrt{m_f}})\right)$$
The error term in the inner sum contributes to
$$\ll \sum_{m_f\leq x} \frac{1}{m_f} \sqrt x$$
Use the first remark about power full number to deduce that this contributes to
$$\ll \sqrt x$$
The error term comes from $\sum_{m_f > x}$ in the main term of inner sum, contributes to 
$$\ll \sum_{m_f > x} \frac{1}{m_f^{3/2}}\ll \sqrt x$$
Hence, we are left with 
$$
x\sum_{m_f }\frac{k(m_f)}{m_f}\left(\frac{6}{\pi^2}\prod_{p|m_f}\left(1+\frac 1 p\right)^{-1}\frac{1}{m_f}\right)+O(\sqrt x)$$
Using Euler product, the sum above becomes 
$$C=\prod_p \left(1-\frac{1}{p(p+1)}\right)$$
Therefore, 
$$
\sum_{n\leq x}\frac{k(n)}{n}= Cx+O(\sqrt x)$$
