# Convergence test

1. Jan 28, 2016

### arpon

1. The problem statement, all variables and given/known data
$P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ...$
Determine if the series is convergent or divergent if $|z| = 2$, where, $z$ is a complex number.

2. Relevant equations
$1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}$

3. The attempt at a solution
Let, $z = 2 exp (i \theta)$

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 $\frac{1}{1+\frac{z}{2}}$

Last edited: Jan 28, 2016
2. Jan 28, 2016

### Samy_A

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.

Last edited: Jan 28, 2016
3. Jan 28, 2016

### arpon

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.

4. Jan 28, 2016

### Samy_A

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.

5. Jan 28, 2016

### PeroK

$\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)$

6. Jan 28, 2016

### Samy_A

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 $|z|=2$.

7. Jan 28, 2016

### PeroK

It doesn't. I thought you'd proved that.

8. Jan 28, 2016

### Samy_A

I thought so too, but given the reputation of the book I started to fear I was missing something. Thanks for the second opinion.

It's not you lacking some important concept here. It surely is a reliable book, but on this point you were right and they made a mistake. Happens to the best.