$f(n)$ is the number of groups of order $n$. Prove that $3 \leq n \implies \frac{n}{\ln(n)} < \sum_{k=1}^{n}\frac{1}{f(k)}$

Question

$f(n)$ is number of groups of order $n$ up to isomorfism. Prove $$3 \leq n \implies \frac{n}{\ln(n)} < \sum_{k=1}^{n}\frac{1}{f(k)}$$

Context:

I am studing this function. Searching for interesting properties. Nobody knows an efficient method to find the number of groups in general. I already know that $f(n)$ cannot be expressed as a finite degree polynomial function.

Let $p$ a prime number. Let $m$ a positive integer. I already know that:

$$p^{\frac{2}{27}m^{2}(m-6)} \leq f(p^{m})$$

Then $f(n)$ grows really fast for some cases. I was wondering if

$$\sum_{k=1}^{\infty} \frac{1}{f(k)}$$

converges. I thought that to search a lower bound would help me.

Current Progress: This is a solved problem.

Answer 1

Let's define functions: $$g_{0}(n) := \frac{1}{f(n)}$$ $$g_{1}(n) := \sum_{k=1}^{n}g_{0}(k)$$ $$g_{2}(n) := \left\{ \begin{array}{@{}ll@{}} 1, & \text{if $n$ is a prime number} \\ 0, & \text{otherwise} \end{array} \right.$$ $$g_{3}(n) := \sum_{k=1}^{n}g_{2}(k)$$

$$g_{2}(n) \leq g_{0}(n) \implies g_{3}(n) \leq g_{1}(n)$$

$g_{3}$ is the prime counting function. For $17 \leq n$:

$$\frac{n}{\ln(n)} < g_{3}(n) \leq g_{1}(n) \implies \frac{n}{\ln(n)} < g_{1}(n)$$

The inequality is also true for $3 \leq n \leq 16$.