# What is $\varlimsup \frac{\omega(n)}{\log n}$?

$\omega(n)$ is the number of distinct prime divisors of $n$. How to figure out?

$$\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}$$

or $\dfrac{\omega(n)}{\log n}$ is convergent, so $\lim\limits_{n\to\infty} \frac{\omega(n)}{\log n}=0$?

It is obvious that $2^{\omega(n)}\leq n$, or

$$\omega(n)\leq \frac{\log n}{\log2}$$

so $\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}\leq\frac1{\log 2}$

Continue to move forward, $n\geq7$, then $\omega(n)\leq \log n$. so

$$\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}\leq1$$

P.s. I think about this problem after reading To show: $\omega(n)\ne\pi(n) , \forall n>2$

#### Solutions Collecting From Web of "What is $\varlimsup \frac{\omega(n)}{\log n}$?"

The largest value of this for each $\omega(n)$ is the primorial with that number of prime factors. With $\theta(x)$ Chebyshev’s first function, you are asking about
$$\limsup \frac{\pi(x)}{\theta(x)} \approx \frac{x}{\log x \; \; x} \rightarrow 0.$$