1. ## Laurent Series-Complex

If anyone could give me a hint that would be great.

Determine the annulus of convergence of the Laurent series and find the sum fo the series there.

Σ n(-1)^n 2^ -│n│Z^n
( sum of n = neg. infinity to infinity)

Thanks

2. Originally Posted by taypez
If anyone could give me a hint that would be great.

Determine the annulus of convergence of the Laurent series and find the sum fo the series there.

Σ n(-1)^n 2^ -│n│Z^n
( sum of n = neg. infinity to infinity)

Thanks
You have,
$\sum_{n=-\infty}^{\infty} \frac{n(-1)^n}{2^{|n|}} z^n$
Necessay and Sufficient (slip the sums),
$\sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{|n|}}z^n$+ $\sum_{n=1}^{\infty} \frac{n(-1)^n}{2^{|n|}}z^n$
You can get rid of the absolute value because it takes on positives and negatives,
$\sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{-n}}z^n$+ $\sum_{n=1}^{\infty} \frac{n(-1)^n}{2^n}z^n$
In the first summand, let $u=1/z$ (if $z=0$ then it converges, nothing is lost).
Thus, we can write,
$\sum_{n=0}^{\infty}\frac{n(-1)^{n+1}}{2^n}u^n+\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^n}z^n$
The non-alternating term is,
$\frac{n}{2^n}$.
Use the generalized ratio test,
$\lim_{n\to \infty}\frac{n+1}{2^{n+1}}\cdot \frac{2^n}{n}=\frac{1}{2}$.
Hence the reciprocal of this, which is 2, is the radius of absolute convergence.
Thus, (in both series)
$|z|<2$
$|u|<2$
Take reciprocal of both sides on the second one,
$1/|u|=|z|>1/2$
Thus,
$1/2<|z|<2$
This is an annulus of absolute convergence.
I still think you need to check the boundary for convergence.

This is my 38th Post!!!

3. Originally Posted by ThePerfectHacker
You have,
$\sum_{n=-\infty}^{\infty} \frac{n(-1)^n}{2^{|n|}} z^n$
Necessay and Sufficient (slip the sums),
$\sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{|n|}}z^n$+ $\sum_{n=1}^{\infty} \frac{n(-1)^n}{2^{|n|}}z^n$
These two series can be summed as they are the derivatives of geometric series which converge for $|z|>1/2$ and $|z|<2$ respectively, and have singularities on the circles $|z|=1/2$ and $|z|=2$ respectively, which is sufficient to conclude that the annulus of convergence is $1/2<|z|<2$.

RonL

4. Originally Posted by CaptainBlack
These two series can be summed as they are the derivatives of geometric series which converge for $|z|>1/2$ and $|z|<2$ respectively, and have singularities on the circles $|z|=1/2$ and $|z|=2$ respectively, which is sufficient to conclude that the annulus of convergence is $1/2<|z|<2$.

RonL
I was thinking about something else, these are alternating, decreasing sequences to zero. Thus, can we use the Leibniz alternating series test? I was not certain about that because this was a complex analysis type question.