Consider the series below where \(r\) is in the interval \((1,-1)\):
\[S = \sum_{i=1}^n ir^i \]
The sum of this series is:
\[S = \frac{r(1-r^n)}{(1-r)^2} - \frac{n}{1-r} r^{n+1}\]
Now let \(n\) go to infinity:
\[\begin{align} S &= \lim_{n \to \infty} \frac{r(1-r^n)}{(1-r)^2} - \frac{n}{1-r} r^{n+1} \\ S &= \lim_{n \to \infty} \frac{r(1-r^n)}{(1-r)^2} - \lim_{n \to \infty} \frac{n}{1-r} r^{n+1} \end{align}\]
This means that when \(n\) goes to infinity, \(r^n\) will go to 0. Therefore:
\[\begin{align} S &= \frac{r(1-0)}{(1-r)^2} - \lim_{n \to \infty} \frac{n}{1-r} r^{n+1} \\ &= \frac{r}{(1-r)^2} - \lim_{n \to \infty} \frac{n}{1-r} r^{n+1} \end{align}\]
What about the second term?
\[\begin{gather} \lim_{n \to \infty} \frac{n}{1-r} r^{n+1} \\ \lim_{n \to \infty} \frac{r}{1-r} nr^n = \frac{r}{1-r} \lim_{n \to \infty} nr^n \end{gather}\]
When \(n\) goes to infinity, \(nr^n\) gives us an indeterminate form \(\infty * 0\), but we can change this to \(\infty/\infty\) form:
\[ \lim_{n \to \infty} nr^n = \lim_{n \to \infty} \frac{n}{r^{-n}}\]
We can use L'Hôpital's rule here:
\[\lim_{n \to \infty} \frac{n}{r^{-n}} = \lim_{n \to \infty} \frac{\frac{d}{dn} n}{ \frac{d}{dn} \frac{1}{r^n} } = \lim_{n \to \infty} \frac{1}{ \frac{-r^n\ln(r)}{r^{2n}} } = \lim_{n \to \infty} \frac{r^{2n}}{-r^n\ln(r)} = \lim_{n \to \infty} \frac{r^n}{-\ln(r)} =0\]
This means:
\[\begin{align} S &= \frac{r}{(1-r)^2} - \lim_{n \to \infty} \frac{r}{1-r} nr^n \\ S &= \frac{r}{(1-r)^2} - 0= \frac{r}{(1-r)^2}\end{align}\]