Question

Determine where sum 1/1+z^n converges in C.

Answer 1

Converge of this?
\[\sum_n \frac{1}{1+z^n}\]

Answer 2

yup

Answer 3

What converge theorems do you have? The ratio test/condition?

Answer 4

Yeah

Answer 5

I assume all of the major ones.

Answer 6

I "know" that it converges for \(\lvert z\rvert > 1\) and diverges otherwise, but I don't know how to show it.

Answer 7

That definitely feels intuitively right. Let a_n be the nth term. Then
\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{1+z^{n+1}}{1+z^n} \right|\]
we want to some show that this ratio, call it R satisfies
\[R \leq \delta < 1\]
for some delta, for sufficiently large x if |z| > 1. Then we will have convergence.

Answer 8

ok

Answer 9

Actually the bounding by delta is probably too strong. It'll be enough that the limit of R = r < 1 if |z| > 1

Answer 10

Whoops, I've written R incorrectly

Answer 11

Now, it seems like it will be much harder in \(\mathbb{C}\).

Answer 12

it's the reciprocal of that. And now I think it's easy.

Answer 13

| a(n+1)/a(n) | = | ( 1 + z^n) / (1 + z^(n+1) ) |
= | ( 1/z^n + 1) / ( 1/z^n + z ) |
Now if |z| > 1, then you can see this limit goes to --> 1/|z| --> 0

Answer 14

It's clear that if |z| =1, for example an irrational point on the unit circle, you're in trouble; and on a rational point it's like to be not even defined for some n.

Answer 15

and inside the unit circle, |1/(1+z^n)| doesn't even go to zero, so the sum certainly can't converge.

Answer 16

right, that part is easy

Answer 17

okay, i think I buy the limit

Answer 18

okay, cool

Answer 19

actually, the best thing to do is divide top and bottom by z^(n+1) then you can break up the | | of the numerator and denominator and show the num --> 0 even if the dem is finite and that will do it.

Answer 20

ok, cool