# An asymptotic property of the divisor counting function

Let $\tau(n)$ be the number of positive divisors of the positive integer $n$ and let $\phi(n)$ be Euler’s function. Then it is easily seen that

$\phi(n)+\tau(n)\le n+1$

for each $n$. So

$\displaystyle\frac{\phi(n)}{n}+\frac{\tau(n)}{n}\le 1+\frac 1n$

holds for all $n$.

$\displaystyle\limsup_{n\to\infty}\frac{\phi(n)}{n}=\lim_{n\to\infty}\frac{n-1}{n}=1$.

So it follows that

$\displaystyle\limsup_{n\to\infty}\frac{\tau(n)}{n}=0$.

Therefore, by the sandwich theorem,

$\displaystyle\lim_{n\to\infty}\frac{\tau(n)}{n}=0$,

i.e., $\tau(n)=o(n)$ as $n\to\infty$. In layman’s terms, this is saying that the number of divisors of $n$ is small compared to $n$ when $n$ is large. The following plot of $\tau(n)/n$ against $n$ for $1\le n\le 500$ captures this nicely.

So this means, for example, that there are only finitely many positive integers $n$ with more than $n/2015$ divisors, which I think is pretty neat!