show that $(n/a_n)$ takes every positive integer value 
Suppose $(a_n)$ is a nondecreasing sequence of positive integers with $\lim\limits_{n\to\infty} \dfrac{a_n}n=0$. Show that $(n/a_n)$ takes every positive integer value.

It turns out that the above implies the following: for every positive integer k, there is an integer N so that there are exactly $N$ primes less than $kN.$

Edit: The latter claim follows since $\pi(n)$ is a nondecreasing sequence of integers that is asymptotically logarithmic in $n$ (though this is very nontrivial to prove)
$$\lim_n\frac{\pi(n)}{n}=0$$
implies that $n/\pi(n)$ takes any positive integer value.

I'm not sure how to prove the second fact. Let $k$ be a positive integer and choose $n$ with $n/a_n = k$. Suppose for a contradiction that there does not exist an integer $N$ so that there are exactly $N$ primes less than $kN$.
Let $m$ be a positive integer. We want to find n with $n/a_n = m\Leftrightarrow 1/m = a_n/n$. It might be useful to consider the finite set $\{k : \dfrac{a_{mk}}{mk}\ge 1/m\}$ for each positive integer m. It is finite since $\lim\limits_{n\to\infty} \dfrac{a_n}n = 0$. $a_n$ is nondecreasing. If $a_n$ is bounded, then since it is nondecreasing, it is eventually constant and equal to say, $d$. Then for large enough k $kd/a_k = k$ so all integers $\ge k$ can be attained.
For integers $<k$, pick the first $n$ so that $n \ge a_n$; there must exist such an n since $\lim\limits_{n\to\infty} a_n/n = 0.$ Then note that $n = a_n$ as if $n > a_n$, the only way $n-1 \leq a_{n-1} = a_n - s< n-s$ for some $s\ge 0$ is if $s < 0,$ a contradiction. So $1$ is definitely attained.
In light of the above, we consider the smallest $k$ so that $\dfrac{a_{mk}}{mk} \ge \dfrac{1}m.$ If $a_{mk} > k,$ then $a_{mk-k} = a_{mk} - s$ for some $s\ge 0$ and by minimality, $k = 1$ or $a_{m(k-1)} \leq k-1$. The latter implies $k-1 \leq a_{m(k-1)} < k-s$, which is clearly impossible. Hence the former must occur, so $a_m > 1.$ This doesn't seem to lead to a contradiction though.
 A: This is easier than it seems. Since
$$\lim_n\frac{\pi(n)}{n}=0,$$
we know that $n/\pi(n)$ takes any positive integer value.
A: For the first part, you have the right general idea but, for example, you should consider each index value instead of just multiples of your $m$. In particular, similar to what you did, for any positive integer $k$, define the set
$$S = \left\{j_k : \frac{a_{j_k}}{j_k}\ge \frac{1}{k}\right\} \tag{1}\label{eq1A}$$
Note it's non-empty since $\frac{a_1}{1}=a_1\ge 1\ge\frac{1}{k}$, so $1 \in S$. Also, since $\lim_{n\to\infty}\frac{a_n}{n} = 0$, then $S$ is finite. Let $j$ be the largest element of $S$. If
$$\frac{a_{j}}{j} = \frac{1}{k} \; \; \to \; \; \frac{j}{a_{j}} = k \tag{2}\label{eq2A}$$
then we're done. Otherwise, the fraction with that index $j$ must be larger than $\frac{1}{k}$ and, since $j+1 \not\in S$, that other fraction must be less than $\frac{1}{k}$. Thus, we have
$$\frac{a_{j}}{j} \gt \frac{1}{k} \; \to \; ka_{j} \gt j \tag{3}\label{eq3A}$$
$$\frac{a_{j+1}}{j+1} \lt \frac{1}{k} \; \to \; ka_{j+1} \lt j+1 \tag{4}\label{eq4A}$$
Since $a_n$ is non-decreasing, then $a_{j+1} \ge a_{j}$. We then get from \eqref{eq3A} and \eqref{eq4A} that
$$j \lt ka_j \le ka_{j+1} \lt j + 1 \tag{5}\label{eq5A}$$
which is not possible for integers. This means the only possibility is that \eqref{eq2A} is true.
