# 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. – GEdgar Aug 22 '16 at 17:08
• It's true if $|r|<1|$ and in fact $\left| \dfrac i 6 \right| < 1.\qquad$ – Michael Hardy Aug 22 '16 at 17:09
• Three answers appear so far, but I'm the only one who's up-voted the question. $\qquad$ – Michael Hardy 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. – Lubin 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. – Pratyush Yadav Aug 23 '16 at 6:45

## 3 Answers

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. – DonAntonio Aug 22 '16 at 17:24

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. – Martin Argerami 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 :( – Polygon Aug 26 '16 at 2:33
• I have edited the answer, you should be able to change your vote if you want to. – Martin Argerami Aug 26 '16 at 4:23

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.