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$\sum_{k \geq 0} \frac{z^k}{z^k+1}$ on the domain $\overline{D}[0, r]$, where $0 \leq r < 1$

I'm honestly not sure how to do this. My text mentions the Weierstrass M-test but the example they gave after stating it uses a completely different method (looks like a repeat of a previous example) and looks nothing like the M-test.

Wolfram Hicks
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2 Answers2

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For $|z|\in [0,r]$, with $0\le r<1$, we have

$$\left|\frac{z^k}{1+z^k}\right|\le \frac{r^k}{1-r}$$

and $\sum_{k=0}^\infty r^k=\frac{1}{1-r}<\infty$

Mark Viola
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From the notation and complex analysis tag I would guess we should assume $z\in \mathbb C,|z|\le r.$ For $k>0,$ we can then say $|1+z^k|\ge 1-|z^k| = 1-|z|^k \ge 1- |z| \ge 1-r.$ Thus

$$\left |\frac{z^k}{1+z^k}\right | = \frac{|z^k|}{|1+z^k|} \le \frac{r^k}{1-r}.$$

Since $\sum_{k=1}^{\infty} r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|\le r$ by Weierstrass M.

zhw.
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  • zhw., do you disagree with user1337? https://math.stackexchange.com/questions/2870028 – BCLC Aug 08 '18 at 12:51
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    @BCLC user1337 has a valid point. Your bound on $$\left |\frac{z^k}{1+z^k}\right |$$ was not correct. – zhw. Aug 08 '18 at 15:15
  • Right right, but I meant to ask about the $\frac{r^k}{1-r^k}$. Anyhoo, Mark Viola answered. Thanks anyhoo! – BCLC Aug 08 '18 at 15:17