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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.

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    $\begingroup$ 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. $\endgroup$ Oct 30, 2019 at 9:20
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    $\begingroup$ Look up Selberg-Delange. $\endgroup$
    – Lucia
    Oct 30, 2019 at 9:36
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    $\begingroup$ 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 $\sum_{n\le x} \frac{n}{\tau(n)} = \sum_{1\le k \le K} c_k x \log^{1/2-k}(x) + O(x \log^{1/2-k}(x))$ $\endgroup$
    – reuns
    Oct 30, 2019 at 14:58

1 Answer 1

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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.

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