Let $\tau(n)$ be the number of positive divisors of $n\in \mathbb{N}$.  

Is it possible to get some good estimate for the sum $\sum\_{n\le x} \frac{n}{\tau(n)}$? 


I know that the sum is $\mathcal O(x^2)$ but I was hoping for something better for example $\sum\_{n\le x} \frac{n}{\tau(n)}=c\cdot x^2+\mathcal O(x^{small power})$.  

Of course this may not be true, but I think it is a safe guess.  

I tried partial summation but it did not help. 



How to estimate the sum $\sum_{n\le x} \frac{n}{\tau(n)}$?
It was proven by Ivic that $$
\sum\_{n \leq x} \frac{1}{d\_{k}(n)}=b\_{k, 1} x \log ^{1 / k-1} x+\cdots+b\_{k, N} x \log ^{1 / k-N} x+O\left(x \log ^{1 / k-N-1} x\right)$$ where $k \geq 2$ and $N$ is arbitrary, fixed, natural number; the constants $b\_{k, 1}, \ldots, b\_{k, N}$ depend only on $ k$ (see *A. Ivic. On the asymptotic formulae for some functions connected with powers of the zeta-function. Mat. Vesnik (Belgrade) 1 (14) (29) (1977), 79–90.*)
Summation by parts transform this result into your sum. 


For $k = 2$ this was stated without proof by Ramanujan and proved in *B.M. Wilson. Proofs of some formulae enunciated by Ramanujan. Proc. London Math. Soc. (2) 21 (1922)
235–255.*


