5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For convergence of the geometric series, the condition is $|r|<1$, where $|\cdot|$ may be taken in the complex sense. $\endgroup$ – GEdgar Aug 22 '16 at 17:08
  • 3
    $\begingroup$ It's true if $|r|<1|$ and in fact $\left| \dfrac i 6 \right| < 1.\qquad$ $\endgroup$ – Michael Hardy Aug 22 '16 at 17:09
  • $\begingroup$ Three answers appear so far, but I'm the only one who's up-voted the question. $\qquad$ $\endgroup$ – Michael Hardy Aug 22 '16 at 17:11
  • $\begingroup$ 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. $\endgroup$ – Lubin Aug 22 '16 at 17:32
  • $\begingroup$ @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. $\endgroup$ – Pratyush Yadav Aug 23 '16 at 6:45
3
$\begingroup$

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:

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

$\endgroup$
  • 2
    $\begingroup$ I think you may have confused the question, which was about the given sequence, not a series. $\endgroup$ – DonAntonio Aug 22 '16 at 17:24
5
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I'm really curious about the reason for the downvote. $\endgroup$ – Martin Argerami Aug 23 '16 at 1:54
  • $\begingroup$ 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 :( $\endgroup$ – Polygon Aug 26 '16 at 2:33
  • $\begingroup$ I have edited the answer, you should be able to change your vote if you want to. $\endgroup$ – Martin Argerami Aug 26 '16 at 4:23
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.