# Dirichlet Test

• Sep 11th 2010, 05:46 AM
lpd
Dirichlet Test
Can some one help me with this question?

Use the Dirichlet test to show that the series $\displaystyle\sum_{n=1}^{\infty}\frac{z^n}{n}
$
.

converves for all $z$ with $|z|=1$ but $z \not=1$.

I know that the Dirichlet test for the convergence of a series of complex numbers can be stated as follows:
Suppose that $a_n$ is a sequence of positive real numbers with $a_n>a_{n+1}$ for all $n$ and $\displaystyle\lim_{n\to\infty}a_n$. Suppose that $b_n$ is a sequence of complex numbers so that there exists $M$ so that, for all $N$, |\sum_{n=1}^{\infty}b_n |<M, THen the series $\sum_{n=1}^{\infty}a_n b_n$ converges.

So how do I start? Just really confused. Do I let $b_n= z^n$ and start from there?
• Sep 11th 2010, 07:09 AM
Opalg
Quote:

Originally Posted by lpd
Can some one help me with this question?

Use the Dirichlet test to show that the series $\displaystyle\sum_{n=1}^{\infty}\frac{z^n}{n}
$
.

converges for all $z$ with $|z|=1$ but $z \not=1$.

I know that the Dirichlet test for the convergence of a series of complex numbers can be stated as follows:
Suppose that $a_n$ is a sequence of positive real numbers with $a_n>a_{n+1}$ for all $n$ and $\displaystyle\lim_{n\to\infty}a_n$ = 0. Suppose that $b_n$ is a sequence of complex numbers so that there exists $M$ so that, for all $N$, $|\sum_{n=1}^{\infty}b_n |. (Upper limit in that sum should be N, not infty.) Then the series $\sum_{n=1}^{\infty}a_n b_n$ converges.

So how do I start? Just really confused. Do I let $b_n= z^n$ and start from there? Yes !

Notice that $\sum_{n=1}^{N}z^n$ is a geometric series with bounded sum, for any given z with |z| = 1, except z = 1. Also, the numbers $a_n=\tfrac1n$ form a decreasing sequence of positive real numbers with limit 0.
• Sep 11th 2010, 07:11 AM
chisigma
According to...

Dirichlet's Test -- from Wolfram MathWorld

... if $a_{n}$ is a sequence of real numbers with $a_{n}\ge a_{n+1} > 0$ and $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and $b_{n}$ a sequence of complex numbers for which $\forall N$ is $\displaystyle |\sum_{n=1}^{N} b_{n}|< M$, being M independent from N, then the series $\displaystyle \sum_{n=1}^{\infty} a_{n}\ b_{n}$ converges. Here is $a_{n} = \frac{1}{n}$ and $b_{n} = e^{i\ n\ \theta}$ , $\theta \ne 2\ k\ \pi$...

Kind regards

$\chi$ $\sigma$
• Sep 14th 2010, 10:58 AM
lpd
How do I determine what "M" is?

Do I need to check what happens at the endpoints?
• Sep 14th 2010, 11:28 AM
Opalg
Quote:

Originally Posted by lpd
How do I determine what "M" is?

You need to show that the sums $\sum_{n=1}^{N}z^n$ have a bound independent of N. This is a geometric series with sum $\tfrac{z(1-z^N)}{1-z}$. Since |z| = 1, it follows that $\Bigl|\sum_{n=1}^{N}z^n\Bigr| = \tfrac{|z||1-z^N|}{|1-z|} \leqslant \tfrac2{|1-z|}.$ So you can take $M = \tfrac2{|1-z|}$.

Quote:

Originally Posted by lpd
Do I need to check what happens at the endpoints?

What endpoints?
• Sep 15th 2010, 01:40 PM
lpd
endpoints. as in checking if z=1 in the series, it will diverge right? becuase the series then becomes a harmonic series, and harmonic series diverges.

thanks for ur help!
• Sep 16th 2010, 12:02 AM
Opalg
Quote:

Originally Posted by lpd
endpoints. as in checking if z=1 in the series, it will diverge right? becuase the series then becomes a harmonic series, and harmonic series diverges.

The way I read the original question, it only asked for values of z with |z|=1 but z not equal to 1. So you don't need to consider the case z=1. But you're right, in that case the series becomes a harmonic series, which diverges.