# How to estimate the sum $\sum_{n\le x} \frac{n}{\tau(n)}$?

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.

• The density of products of few primes is 0, so your sum is $o(x^2)$. On the other hand, it is (asymptotically) no smaller than $cx^2/\log x$, just by looking at primes. Oct 30, 2019 at 9:20
• Look up Selberg-Delange. Oct 30, 2019 at 9:36
• The Euler product is $\sum_n n^{1-s}/\tau(n) = \prod_p (1+\sum_m p^m/(m+1) p^{-sm})= \zeta(s-1)^{1/2} F(s)$ with $F$ an Euler product converging absolutely on a larger domain thus from the Tauberian theorems the summatory function satisfies $\sum_{n\le x} \frac{n}{\tau(n)} = \sum_{1\le k \le K} c_k x \log^{1/2-k}(x) + O(x \log^{1/2-k}(x))$ Oct 30, 2019 at 14:58

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.