Convergence or divergence

May 2009
86
1
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
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle z\) a complex number?

If so, \(\displaystyle z = r\cos{\theta} + i\,r\sin{\theta}\)

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


So \(\displaystyle a_n = \frac{1}{1 + z^n}\)

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

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

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

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

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


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

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


Now investigate the behaviour of both of these sums.