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Thread: Dirichlet Test

  1. #1
    lpd
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    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} <br />
.

    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?
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  2. #2
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    Quote Originally Posted by lpd View Post
    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} <br />
.

    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  |<M. (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.
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  3. #3
    MHF Contributor chisigma's Avatar
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    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
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  4. #4
    lpd
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    How do I determine what "M" is?

    Do I need to check what happens at the endpoints?
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  5. #5
    MHF Contributor
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    Quote Originally Posted by lpd View Post
    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 View Post
    Do I need to check what happens at the endpoints?
    What endpoints?
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  6. #6
    lpd
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    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!
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  7. #7
    MHF Contributor
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    Quote Originally Posted by lpd View Post
    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.
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