# Thread: convergence of a power series

1. ## convergence of a power series

I have the following problem: Prove that the power series \sum \frac {z^n}{n} converges at every point of the unit circle except z=1.

The part with z=1 I already did it, but I have problems with the rest. Is there a way to prove the convergence using Fourier coefficients?

Thanks.

2. ## Re: convergence of a power series

Let $\displaystyle s_j:=\sum_{k=0}^jz^k$. We have for $\displaystyle n\geq 1$:
\displaystyle \begin{align*}\sum_{k=1}^n\frac{z^k}k&=\sum_{k=1}^ n\frac{s_k-s_{k-1}}k\\&=\sum_{j=1}^n\frac{s_j}j-\sum_{j=0}^{n-1}\frac{s_j}{j+1}\\ &=\frac{s_n}n+\sum_{j=1}^ns_j\left(\frac 1j-\frac 1{j+1}\right)\end{align*}
Now show that for all integer $\displaystyle j$ we have $\displaystyle |s_j|\leq \frac 1{\sin\left(\frac{\theta}2\right)}$.

3. ## Re: convergence of a power series

Originally Posted by Veve
I have the following problem: Prove that the power series \sum \frac {z^n}{n} converges at every point of the unit circle except z=1.

The part with z=1 I already did it, but I have problems with the rest. Is there a way to prove the convergence using Fourier coefficients?

Thanks.
The Abel test for convergence of a series extablishes that...

a) given a sequence $\displaystyle a_{n}$ such that $\displaystyle \forall k$ the partial sum $\displaystyle s_{k}= \sum_{n=1}^{k} a_{n}$ is bounded...

b) given a sequence $\displaystyle b_{n}$ so that $\displaystyle |b_{n}|$ is decreasing and $\displaystyle \lim_{n \rightarrow \infty} b_{n}=0$...

... then the series $\displaystyle \sum_{n=1}^{\infty} a_{n}\ b_{n}$ converges. In Your case $\displaystyle b_{n}=\frac{1}{n}$ satisfies to criterion b), so that we have to verify thwe criterion a) for the sequence $\displaystyle a_{n}= e^{i n \theta}\, \ \theta \ne 0$. The partial sum is...

$\displaystyle s_{k}= \sum_{n=1}^{k} e^{i n \theta} = e^{i \theta}\ \frac{1-e^{i k \theta}}{1-e^{i \theta}}$ (1)

... so that is...

$\displaystyle |s_{k}| = \sqrt{\frac{1-\cos k \theta}{1-\cos \theta}}$ (2)

... and for $\displaystyle \theta \ne 0$ it is evident that (2) is bounded. Then for Abel's test the series $\displaystyle \sum_{n=1}^{\infty} \frac{e^{i \theta}}{n}$ converges if $\displaystyle \theta \ne 0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$