1. convergenze

is $sum(nz^z)$ convergent?

2. 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

$\chi$ $\sigma$

3. Originally Posted by chisigma
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.

4. 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

$\chi$ $\sigma$

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 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

$\chi$ $\sigma$