Results 1 to 5 of 5

Math Help - convergenze

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    1

    convergenze

    is sum(nz^z) convergent?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If you intend...

    \sum_{n}n\cdot z^{z} = z^{z} \cdot \sum_{n} n

    ... it diverges \forall z ...

    If you intend however ...

    \sum_{n} z^{n\cdot z}

    ... that's a nice problem ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    Quote Originally Posted by chisigma View Post
    If you intend however ...

    \sum_{n} z^{n\cdot z}

    ... that's a nice problem ...

    Kind regards

    \chi \sigma
    Just looking at this I would guess it converges for |z|<1 since for large n it will look like a geometric series. I would try using the ratio/root test to prove this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If we write the series as ...

    \varphi(z)= \sum_{n=0}^{\infty} (z^{z})^{n} (1)

    ... it is evident that (1) is a true geometric series so that it converges for...

    |z^{z}|<1 (2)

    At this point the question is: where in the domain of z (2) is verified? ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Remembering the identity...

    z^{z}= e^{z\cdot \ln z} (1)

    ... the condition for which the sum...

    \varphi(z)= \sum_{n=0}^{\infty} z^{n\cdot z} (2)

    ... converges becomes...

    |z^{z}|<1 \rightarrow \Re(z\cdot \ln z)<0 (3)

    If z is a pure real variable, then z^{z}= x^{x}= e^{x\cdot \ln x}, i.e. the function you see here...






    It is remarkable the fact that x^{x} is real for x>0 and complex [with real and imaginary part...] for x<0. In this case the condition (3) is satisfied for 0<x<1 and x<-1 and here the (2) is...

    \varphi(x)= \sum_{n=0}^{\infty} x^{n\cdot x}= \frac{1}{1-x^{x}} (4)

    But what does it happen when z has an imaginary part that is not 0?... that's the problem ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum