What is $\sum_{n=1}^{\infty}\left(\frac i6\right)^n$ where $i=\sqrt{-1}$? What is $\sum_{n=1}^{\infty}\left(\frac i6\right)^n$? where $i=\sqrt{-1}$
I want to evaluate the GP
$$\frac i6 + \left(\frac i6\right)^2+\cdots \infty$$
I am thinking about using the formula for an infinite GP in reals: $\frac{a}{1-r}$
This is true if $r\lt 1$. But the comparison $\frac i6 \lt 1$ is invalid.
So, I went to the initial formula, sum $= \frac{a(1-r^n)}{1-r}$. Our above formula will be valid if $\lim_{n\to\infty}\left(\frac i6\right)^n$ is $0$.
That's where I'm stuck. I am not sure how to evaluate this limit.

Look at, when $|x|<1$:
$$\sum_{n=\text{a}}^{\infty}x^n=\frac{x^{\text{a}}}{1-x}$$
By the geometric series test, the series diverges.
So, when $x=\frac{i}{6}$, check if the condition is true:
$$\left|\frac{i}{6}\right|=\frac{\left|i\right|}{\left|6\right|}=\frac{1}{6}<1$$
So:
$$\text{S}_0=\sum_{n=0}^{\infty}\left(\frac{i}{6}\right)^n=\frac{1}{1-\frac{i}{6}}=\frac{36+6i}{37}$$
$$\text{S}_1=\sum_{n=1}^{\infty}\left(\frac{i}{6}\right)^n=\frac{\frac{i}{6}}{1-\frac{i}{6}}=\frac{-1+6i}{37}$$
