# Convergence test

## Homework Statement

##P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ... ##
Determine if the series is convergent or divergent if ## |z| = 2 ##, where, ## z## is a complex number.

## Homework Equations

##1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}##

## The Attempt at a Solution

Let, ##z = 2 exp (i \theta)##[/B]
For the first ##N## terms, the summation is,
##P_N (z) = \frac{1-(-1)^N exp(iN\theta)}{1+exp(i \theta)}=\frac{1-(-1)^N \cos (N\theta) - i (-1)^N \sin (N \theta)}{1 + exp (i \theta)}##
As ## N \rightarrow \infty##, ##\cos (N \theta)## and ## \sin (N \theta)## do not converge to a particular value.
So, I conclude that the series is not convergent for ## |z|=2##
But the answer says that the series converges to ##\frac{1}{1+\frac{z}{2}}##

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Samy_A
Homework Helper

## Homework Statement

##P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ... ##
Determine if the series is convergent or divergent if ## |z| = 2 ##, where, ## z## is a complex number.

## Homework Equations

##1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}##

## The Attempt at a Solution

Let, ##z = 2 exp (i \theta)##[/B]
For the first ##N## terms, the summation is,
##P_N (z) = \frac{1-(-1)^N exp(iN\theta)}{1+exp(i \theta)}=\frac{1-(-1)^N \cos (N\theta) - i (-1)^N \sin (N \theta)}{1 + exp (i \theta)}##
As ## N \rightarrow \infty##, ##\cos (N \theta)## and ## \sin (N \theta)## do not converge to a particular value.
So, I conclude that the series is not convergent for ## |z|=2##
But the answer says that the series converges to ##\frac{1}{1+\frac{z}{2}}##
The general term of the series is ##a_n=\frac{(-1)^nz^n}{2^n}##.
Then, for ##|z|=2##, ##|a_n|=|\frac{z^n}{2^n}|=\frac{|z|^n}{2^n}=1##. How can this series converge if the sequence ##a_n## doesn't converge to 0?

I think you are correct.

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arpon
The general term of the series is ##a_n=\frac{(-1)^nz^n}{2^n}##.
Then, for ##|z|=2##, ##|a_n|=|\frac{z^n}{2^n}|=\frac{|z|^n}{2^n}=1##. How can this series converge if the sequence ##a_n## doesn't converge to 0?

I think you are correct.
Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.

Samy_A
Homework Helper
Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.
I see. They write that it converges for all ##z## with ##|z|=2## except ##z=-2##.
Let's take ##z=2## then. The series becomes ##1-1+1-1 ...##, so the partial sums alternate between ##0## and ##1##.
A few lines before this they give an example with the same oscillating series and conclude (correctly in my book) that it doesn't converge:

Call me baffled.

PeroK
Homework Helper
Gold Member
##\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)##

arpon
Samy_A
Homework Helper
##\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)##
But the book claims that this series also converges for all ##z## with ##|z|=2## except ##z=-2##. This is clearly a highly regarded book, but still, I can't see how the series converges when ##|z|=2##.

PeroK
Homework Helper
Gold Member
But the book claims that this series also converges for all ##z## with ##|z|=2## except ##z=-2##. This is clearly a highly regarded book, but still, I can't see how the series converges when ##|z|=2##.
It doesn't. I thought you'd proved that.

Samy_A