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Let $$f(z)=\sum_{n=0}^{\infty}\frac{z^{n [\mathbb{Re} z]}}{n}$$

For which $z\in\mathbb{C}\setminus\{0\}$ does this series converge?

I have trouble with this example. When I use the ratio test, I get:

$$|z^{[\mathbb{Re} z]}|<1$$

I can't solve it. Also, I know that it is not enough, because I should also check the case when $$|z^{[\mathbb{Re} z]}|=1$$

Note that $[\mathbb{Re} z]$ means the integer part of $\mathbb{Re} z$.

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  • $\begingroup$ For some $x$ expressed in terms of $z$, you series is $\sum\cfrac{x^n}{n}$ for which you should be able to answer the question. $\endgroup$ – xavierm02 Nov 22 '14 at 17:49
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I think that noticing that for $\forall z \,\,/ \,\, \text{Re}(z) \in (-1,1) \,\, |z^{[\text{Re}(z)]}| = 1$ might be useful as it implies that $|z^{[\text{Re}(z)]}|\ge 1 \forall z$.

EDIT:

Write $z^{[\text{Re}(z)]} =k$. The sum:

\begin{equation} \sum_{i=1}^{+\infty} \frac{k^n}{n} \end{equation}

Has a convergence radius of 1, so absolutely converges when $|k| < 1$. In our case, never.

We may be interested in the case in which $|k|=1$:

  • If $\text{Re}(z) \in (-1,1)$, the series becomes the harmonic series;
  • The same when $\text{Re}(z)=1$ and $Im(z)=0$;
  • When $Re(z)=-1$ and $Im(z)=0$, then the series converges because it is the alternate harmonic series;
  • Elsewhere it cannot converge as $|k|$ would be bigger than 1 by definition of complex modulus.
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  • $\begingroup$ I know, but it isn't enough. $\endgroup$ – luka5z Nov 22 '14 at 17:42

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