Homework Help: 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.