Proportion of numbers with $\tau(n)>\ln{N}$ We find the average value for $\tau(n)$ (number of divisors of $n$) for $1 \leqslant n \leqslant N$. 
$$\text{Number of $n$ with divisor $i$} = \bigg\lfloor\frac{N}{i}\bigg\rfloor$$
$$\sum_{n=1}^N \tau(n)=\sum_{i=1}^N \bigg\lfloor\frac{N}{i}\bigg\rfloor \sim \sum_{i=1}^N \frac{N}{i} \sim N\ln{N}$$
$$\text{Average value of $\tau(n)$ for $1 \leqslant n \leqslant N$} = \frac{1}{N}\sum_{i=1}^N \bigg\lfloor\frac{N}{i}\bigg\rfloor \sim \ln{N}$$
Thus, we would expect a random number chosen from $1$ to $N$ to have around $\ln{N}$ divisors.
Let $f(N)$ be the number of values $1 \leqslant n \leqslant N$ such that $\tau(n) > \ln{N}$. What is the value of:
$$ \lim_{N \to \infty} \frac{f(N)}{N}$$
In other words, what is the proportion of numbers till $N$ that have more than $\ln{N}$ divisors as $N$ tends to $\infty$ ?
 A: Hardy-Ramanujan theorem shows that normally we have $\tau(n)$ is about $2^{\log\log n}$. We have
$$2^{\omega(n)}\leq \tau(n)\leq 2^{\Omega(n)}$$
where $\omega(n)$ is the number of distinct factors of $n$, and $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity.
By Hardy-Ramanujan, both $\omega(n)$ and $\Omega(n)$ are about $\log\log n$ normally. Then we see that $\tau(n)>\log N$ is much larger than the normal order $2^{\log\log n}=(\log n)^{\log 2}$. 
To make this observation more rigorous, we use the following theorem (see Montgomery & Vaughan 'Multiplicative Number Theory I, Classical Theory' Theorem 7.20.

Theorem Let $1\leq r\leq R<2$. Then for sufficiently large $N$, the number of $n\leq N$ satisfying $$\Omega(n)\geq r\log\log N$$
  is $\leq BN (\log N)^{r-1-r\log r}$ for some absolute constant $B>0$. 

If we have $2^{\Omega(n)}\geq \tau(n)> \log N$, then 
$$
\Omega(n)\geq \frac1{\log 2} \log\log N.
$$
Applying $r=1/\log 2$ in the above theorem, we obtain that the number of $n\leq N$ with $\tau(n)>\log N$ is bounded above by 
$$
BN(\log N)^{r-1-r\log r}.
$$
A calculation shows that we have
$$
r-1-r\log r < -0.086.
$$
Therefore, we have for sufficiently large $N$, 
$$
\frac{f(N)}N \leq B (\log N)^{-0.086}.
$$
Hence, we have 
$$
\lim_{N\rightarrow\infty} \frac{f(N)}N = 0.
$$
