# Convergence of power series

• March 28th 2011, 05:30 PM
mulaosmanovicben
Convergence of power series
basic complex analysis (so z is complex):

Prove that the power series $\sum_{k=0}^{\infty} z^{k}$ converges at no point on its circle of convergence |z| = 1

Attempt:

I have no idea what to do. This course is driving me nuts. I know for |z| < 1 that series converges to 1/1-z but that is all I know. Can anyone give me a hint?
• March 28th 2011, 05:40 PM
Prove It
It is well known that geometric series are divergent where $\displaystyle |r| \geq 1$.

If you must prove it though, since $\displaystyle |z| = 1$, then we can write $\displaystyle z = e^{i\theta}$.

So $\displaystyle \sum_{k = 0}^{\infty}z^k = \sum_{k = 0}^{\infty}\left(e^{i\theta}\right)^k$

$\displaystyle = \sum_{k = 0}^{\infty}e^{ik\theta}$

$\displaystyle = \lim_{n \to \infty}\left[\frac{e^{i\theta}\left(e^{i(n+1)\theta} - 1\right)}{e^{i\theta} - 1}\right]$.

Can this limit be evaluted? If so, what does it go to?
• March 28th 2011, 05:42 PM
Drexel28
Quote:

Originally Posted by mulaosmanovicben
basic complex analysis (so z is complex):

Prove that the power series $\sum_{k=0}^{\infty} z^{k}$ converges at no point on its circle of convergence |z| = 1

Attempt:

I have no idea what to do. This course is driving me nuts. I know for |z| < 1 that series converges to 1/1-z but that is all I know. Can anyone give me a hint?

You know that if $|z|=1$ that $\displaystyle z=e^{i\theta}$ so that you're question reduces to $\displaystyle \sum_{k=0}^{\infty}e^{ik\theta}$ but a quick check shows that (assuming $z\ne 1$ but that case is trivial) $\displaystyle \sum_{k=0}^{m}e^{ik\theta}=\frac{1-e^{i(m+1)\theta}}{1-e^{i\theta}}$ and so to assume that $\displaystyle \sum_{k=0}^{\infty}z^k$ converges is to assume that $\displaystyle \lim_{m\to\infty}e^{i(m+1)\theta}$ exists...I think you can take it from here.
• March 28th 2011, 07:14 PM
mulaosmanovicben
Quote:

Originally Posted by Prove It
It is well known that geometric series are divergent where $\displaystyle |r| \geq 1$.

If you must prove it though, since $\displaystyle |z| = 1$, then we can write $\displaystyle z = e^{i\theta}$.

So $\displaystyle \sum_{k = 0}^{\infty}z^k = \sum_{k = 0}^{\infty}\left(e^{i\theta}\right)^k$

$\displaystyle = \sum_{k = 0}^{\infty}e^{ik\theta}$

$\displaystyle = \lim_{n \to \infty}\left[\frac{e^{i\theta}\left(e^{i(n+1)\theta} - 1\right)}{e^{i\theta} - 1}\right]$.

Can this limit be evaluted? If so, what does it go to?

Why is the summation equal to that limit?
• March 28th 2011, 08:13 PM
Prove It
Because for a finite geometric series, the sum of the first $\displaystyle n$ terms is $\displaystyle \frac{a\left(r^n - 1\right)}{r - 1}$.

The infinite sum will be the limit of this as $\displaystyle n \to \infty$.
• March 29th 2011, 01:18 AM
FernandoRevilla
An alternative:

If $|z|=1$ then, $|z^k|=1$ which implies $l=\lim_{k\to +\infty}z^k$ does no exist or $l\neq 0$ . In both cases, the necessary condition for he convergence of $\sum_{k\geq 0}z^k$ is not satisfied.