is $\displaystyle sum(nz^z)$ convergent?

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- May 7th 2009, 03:55 AM #1

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- May 7th 2009, 04:08 AM #2
If you intend...

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

... it diverges $\displaystyle \forall z$ ...

If you intend however ...

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

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

Kind regards

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

- May 8th 2009, 02:24 PM #3

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- May 8th 2009, 11:44 PM #4
If we write the series as ...

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

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

$\displaystyle |z^{z}|<1$ (2)

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

Kind regards

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

- May 9th 2009, 07:19 AM #5
Remembering the identity...

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

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

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

... converges becomes...

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

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

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

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

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

Kind regards

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