Proving $a_n = \frac{q^n}{n}$ diverges

Question

So, I have recently been studying sequences and I'm having quite a bit of trouble with the following problem:

Let $a_n = \frac{q^n}{n}$ with $q > 1, n \in \mathbb{N}$. Find the limit of $a_n$ if it exists.

Now, I know that this is a form of a geometric sequence $(q^n)_{n \in \mathbb{N}}$, which diverge for $|q| > 1$ and plotting the graph of $a_n$ for some values of $q$ shows that $a_n$ will most likely also diverge towards $\infty$. However, I just can't quite prove that this is the case. I attempted to show the following to prove this diverges:

$$\forall K>0: \exists N=N(K) \in \mathbb{N}: \forall n > N: a_n > K$$

The most obvious way to show this would be to show that this sequence is increasing from some point on-wards for all $q > 1$ and then solving $\frac{q^n}{n} > K$ for n. However, this didn't help much since I don't know how to solve that equation and I also feel like there should be some easier or more obvious way to show $a_n$ diverges.

Answer 1

Since $\lim \frac{a_{n+1}}{a_n}=q \gt 1,$ choose $N$ such that $n \gt N \Rightarrow \frac{a_{n+1}}{a_n} \gt \frac {1+q}{2} \gt 1$. From that point forward, the sequence must get and stay larger than a geometric sequence with ratio $\frac{1+q}{2}$, which you know diverges. Since the sequence in question is eventually larger than a sequence that diverges to $\infty$, our sequence also must do so.