Convergence of $r^n/n$ when $|r| > 1$. Consider $r\in \mathbb{R}$ and the real number sequence $(a_n)_{n\in \mathbb{N}}$ where
$$a_n = \dfrac{r^n}{n}.$$
If $|r|<1$, we know that $r_n\to 0$ when $n\to \infty$. Since $1/n \to 0$ when $n\to \infty$ by theorem about sequences we know that since
$$a_n = \left(\dfrac{1}{n}\right)(r^n),$$
we have $a_n\to 0$ when $n\to \infty$.
If, on the other hand $|r|=1$, then we have
$$\left|a_n\right|=\left|\dfrac{r^n}{n}\right|=\dfrac{1}{n},$$
and thus it easily follows that $a_n\to 0$ when $n\to \infty$ because of the archimedean property of the real numbers.
Finally we have the case $|r| >1$. In this case I'm having a hard time. Intuition tells me that the sequence will diverge, but I'm unsure on how to prove this.
What I know is that if $|r|>1$ then the sequence $r^n$ is unbounded, but I couldn't get anything from this. I tried then considering first the case $r >1$ where we know that we can write $r = 1 + h$ so that $r^n \geq 1 + nh$ with $h > 0$, but this got me nowhere.
How can I show that $a_n$ diverges when $|r| > 1$? I'm more interested here in the strategy and how should we think about it rather than just a solution.
 A: Strategy: One term in the series fails because it grows linearly. Two terms grow quadratically, so let's do that.
General idea: Exponential growth always kills polynomial growth - one way to be explicit about this is to look at the Taylor series of $e^x$.

Take one more term in the series expansion. Note that
$$e^{t} \ge 1 + t + \frac{t^2}{2}$$
so that
$$r^n = e^{(\ln r) n} \ge 1 + (\ln r) n + \frac{(\ln r)^2}{2} n^2$$
so that $r^n / n$ is unbounded.
A: You were on the right track.  Let $r=1+\delta$, $\delta>0$.  Then, from the binomial theorem, we can write
$$\begin{align}
r^n&=(1+\delta)^n\\\\
&=1+n\delta +\frac{n(n-1)}{2}\delta^2+\cdot +\delta^n\\\\
&\ge \frac{n(n-1)}{2}\delta^2 \tag 1
\end{align}$$
Therefore, dividing both sides of $(1)$ by $n$ reveals
$$\frac{r^n}{n}\ge \frac12 \delta^2 (n-1)\to \infty \,\,\text{as}\,\,n\to \infty$$
as expected!
A: You can take the limit when $n\to\infty$ of the coefficients
\begin{equation}
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{r^{n+1}}{r^n}\frac{n}{n+1}=r
\end{equation}
So, for big $n$, the coefficients $a_n$ scale as $r^n$, meaning that the convergency of the original sequence will be same as the $a_n=r^n$ sequence.
A: Hint: $r=1+h$ with $h>0$, then show $r^{n}\geq 1+nh+n (n-1)h^{2}/2$
