Prove that $\lim_{n \to \infty}\dfrac{\alpha(n)}{n} = 0$

Question

Prove that $\displaystyle \lim_{n \to \infty}\dfrac{\alpha(n)}{n} = 0$ where $\alpha(n)$ is the number of primes which divide $n$.

I think we should get an upper bound on $\alpha(n)$ by using the fact that each prime is greater than or equal to $2$, but I am not sure who to get the bound. Also, do they mean $\alpha(n)$ to be the number of distinct primes which divide $n$

Answer 1

You are on the right track. Each prime $p$ dividing $n$ must be at least $2$. Hence, there are at most $[\log_{2} n]$ primes dividing $n$, where $[\cdot]$ is the floor function. This assumes that $\alpha(n)$ is the number of primes dividing $n$; if they must be distinct, the upper bound is still valid. Therefore,

$$ \alpha(n)/n\leq\log_{2}(n)/n\to 0 $$ as $n\to\infty$.

Answer 2

Outline: Distinct or not makes no difference to the limit: it is $0$ in either case.

Note that under either interpretation $2^{\alpha(n)}\le n$, so $\alpha(n)\le \log_2(n)=\frac{\ln n}{\ln 2}$. Now use the fact that $n$ goes to infinity much faster than $\ln n$.

Answer 3

The smallest prime that can divide some $n$ is $2$. This means there are at most $\lfloor \log_2(n) \rfloor$ primes dividing $n$.

$$\displaystyle \lim_{n \to \infty}\dfrac{\alpha(n)}{n} \leq \lim_{n \to \infty}\dfrac{\lfloor \log_2(n) \rfloor}{n} = 0$$