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.

You have, since $|i|=1$,
$$
\left|\frac{i^n}{6^n}\right|=\frac1{6^n}.
$$
So the limit is zero. 
