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.

• For convergence of the geometric series, the condition is $|r|<1$, where $|\cdot|$ may be taken in the complex sense. Aug 22 '16 at 17:08
• It's true if $|r|<1|$ and in fact $\left| \dfrac i 6 \right| < 1.\qquad$ Aug 22 '16 at 17:09
• Three answers appear so far, but I'm the only one who's up-voted the question. $\qquad$ Aug 22 '16 at 17:11
• I’ll upvote, with the comment that all you would have needed to do is work out $(i/6)^n$ for a few values of $n$ to see that the limit is zero. If you were worried about convergence in $\Bbb C$, you could have asked about the distance between these numbers and $0$, in the Gaussian plane. If still worried, then you need to read up on the plane and other metric spaces, as geometries in which the concept of “convergence” makes sense. Aug 22 '16 at 17:32
• @Lubin I did try working out $(\frac i6)^n$ for some values of $n$. But, for odd $n$, I was getting either $\frac i{6^n}$ or $\frac {-i}{6^n}$, which I couldn't say was less than $1$. Of course, after reading the answers, I get that I wasn't supposed to be calculating $(\frac i6)^n$, but rather its modulus. I just didn't mention that because I found it irrelevant. Aug 23 '16 at 6:45

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

• I'm really curious about the reason for the downvote. Aug 23 '16 at 1:54
• I'm really sorry man. It looks like I somehow did it on accident. I can't undo it now because it's been too long :( Aug 26 '16 at 2:33
• I have edited the answer, you should be able to change your vote if you want to. Aug 26 '16 at 4:23

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:

$$\text{S}_0=\sum_{n=0}^{\infty}\left(\frac{i}{6}\right)^n=\frac{1}{1-\frac{i}{6}}=\frac{36+6i}{37}$$ $$\text{S}_1=\sum_{n=1}^{\infty}\left(\frac{i}{6}\right)^n=\frac{\frac{i}{6}}{1-\frac{i}{6}}=\frac{-1+6i}{37}$$

• I think you may have confused the question, which was about the given sequence, not a series. Aug 22 '16 at 17:24

You could break the sequence into the four subsequence pieces given by $i$ having order $4$.

Depending on whether $n$ is congruent to $1,2,3,0$ modulo $4$. It will be easy to see that each subsequence converges to 0.