How to find an upper bound for $f(n)=\sum_{k=1}^{n}\frac{1}{d^{9}(k)}$?

Question

How to find an upper bound for

$$f(n)=\sum_{k=1}^{n}\frac{1}{d^{9}(k)}$$

where $d(n)$ is the divisor function?

Answer 1

First, note that $d(k) \ge 2^{\omega(k)}$, where $\omega(k)$ is the number of distinct prime factors of $k$. Therefore $$ f(n) \le \sum_{1\le k\le n} \frac1{(2^{\omega(k)})^9} = 1 + \sum_{m=1}^\infty \frac1{512^m} \sum_{\substack{2\le k\le n \\ \omega(k)=m}} 1. $$ This inner sum has been well studied: a result of Hardy and Ramanujan tells us that there exist positive constants $A$ and $C$ such that $$ \sum_{\substack{2\le k\le n \\ \omega(k)=m}} 1 \le \frac{Ax(\log\log x+C)^{m-1}}{(m-1)!\log x}. $$ Therefore \begin{align*} f(n) &\le 1+ \sum_{m=1}^\infty \frac1{512^m} \frac{An(\log\log n+C)^{m-1}}{(m-1)!\log n} \\ &= 1+ \frac{An}{512\log n} \sum_{m=0}^\infty \frac1{512^m} \frac{(\log\log n+C)^m}{m!} \\ &= 1+ \frac{An}{512\log n} e^{(\log\log n+C)/512} \le \frac{Dn}{(\log n)^{511/512}} \end{align*} for some positive constant $D$.

Moreover, other than the value of the constant $D$, this is the best bound possible: one can show that the contribution of squarefree integers with about $(\log\log n)/512$ prime factors is legitimately of size $n/(\log n)^{511/512}$. If one really wanted to, one could get an asymptotic formula by invoking the Selberg-Sathe formula.

Answer 2

I would start with Dirichlet's theorem $$\sum_{k=1}^n d(n) =n \ln n + (2\gamma-1)n + O(\sqrt{n}) $$ and use the power-means inequality to try to get what you want.