Prove a sequence as 1/k as a cluster point and show

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I'm going through a set of practice problems on my own and I'm stuck on this one:

Let $h(n)$ be the largest prime factor of the integer $n > 1$, and $s(n)$ be the sum of its prime factors, so $h(12) = 3$, $s(12) = 7$.

Prove the sequence $\frac{h(k)}{s(k)}, n = 2, 3, 4, . . .$ has $1/k$ as a cluster point for every positive integer $k$, but no limit.

I think I need to construct a formula for both h(n) and s(n) and then show that 1. $|\frac{h(k)}{s(k)} - \frac{1}{k}| < \varepsilon$ and $\frac{h(k)}{s(k)}$ converges to infinity. However I'm not sure how to do that.

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You do not need to construct a formula for $h(n)$ and $s(n)$. All you need is to find enough values of $k$ for which $\frac{h(k)}{s(k)}$ is close to $\frac1k$.

To do that, think about when $\frac{h(k)}{s(k)}$ is close to $\frac1k$. That will happen if $s(k)$ is "about $k$ times bigger than $h(k)$". To do that, consider what happens when $k=p\cdot 2^n$ for some prime value $p$ and some integer $n$.


Once you prove that each $\frac1k$ is a cluster point of the sequence, it is trivial to argue that the sequence cannot have a limit. You can simply use the fact that convergent sequences, i.e. those with a limit, always have only one cluster point.