# Convergence or divergence

• May 22nd 2010, 04:52 AM
poorna
Convergence or divergence
I am not sure how to work this problem out.

Investigate the behavior (convergence or divergence ) of
∑an if
an = 1/(1 + z^n)
Any help will be appreciated
• May 22nd 2010, 05:45 AM
Prove It
Quote:

Originally Posted by poorna
I am not sure how to work this problem out.

Investigate the behavior (convergence or divergence ) of
∑an if
an = 1/(1 + z^n)
Any help will be appreciated

Is $z$ a complex number?

If so, $z = r\cos{\theta} + i\,r\sin{\theta}$

and $z^n = r^n\cos{n\theta} + i\,r^n\sin{n\theta}$.

So $a_n = \frac{1}{1 + z^n}$

$= \frac{1}{1 + r^n\cos{n\theta} + i\,r^n\sin{n\theta}}$

$= \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{(1 + r^n\cos{n\theta})^2 + r^{2n}\sin^2{n\theta}}$

$= \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}\cos^2{\theta} + r^{2n}\sin^2{n\theta}}$

$= \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}$

$= \frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}} + i \left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)$

So $\sum_n{\left(\frac{1}{1 + z^n}\right)} = \sum_n{\left[ \frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}} + i \left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)\right]}$

$= \sum_n{\left(\frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)} + n\,i\sum_n{\left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)}$.

Now investigate the behaviour of both of these sums.