# When does this complex series converge?

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 $\mathbb{Re} z$.

• 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. – xavierm02 Nov 22 '14 at 17:49

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:

$$\sum_{i=1}^{+\infty} \frac{k^n}{n}$$

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.
• I know, but it isn't enough. – luka5z Nov 22 '14 at 17:42