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The original question is: Show $\sum _{k=0}^{\infty }\frac{z^{k}}{z^{2k}+1} $ is uniformly convergent on $\bar{D}_{r}(0)=\{z:\left | z \right |\leq r \}$ where $0< r< 1$ .

I attempted to use Weierstrass M-test to show it. I want to show that $\forall z\in \bar{D}_{r}(0)$, $\left | \frac{z^{k}}{z^{2k}+1} \right | \leq \left | z^{k} \right |\leq r^{k}=M_{k}$ where $\sum _{k=0}^{\infty }M_{k}$ is a convergent geometric series.

Hence, by Weierstrass M-test, $\sum _{k=0}^{\infty }\frac{z^{k}}{z^{2k}+1}$ is uniformly convergent.

But, I'm getting stuck at showing this part: $\left | \frac{z^{k}}{z^{2k}+1} \right | \leq \left | z^{k} \right |$ ,which is similar to showing $\left | z^{2k}+1 \right |\geq 1 $ .

Can someone please help me?

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    Because it's _not_ true. Instead you need to show for any $00$ such that $|z^{2k}+1|\ge\beta$ for $|z| – 2017-01-02
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    This is false. Take $z=\frac{r}{2}\exp (i \pi /2)$, then, $|z^{2}+1|=1-\frac{r^2}{4}$. – 2017-01-02

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Hint. Use $$ |a+b|\ge||a|-|b||,\quad a,b \in \mathbb{C}, $$ with $a=z^{2k}$, $b=1$, $|z|

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    Hmmm..how? Still can't get it. =( – 2017-01-02
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    @Elina You may write $|z^{2k}+1| \ge ||z|^{2k}-|1||=1-|z|^{2k}>1-r^{2k}>1-r>0$ since $|z| – 2017-01-02
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    @Elina Did you get it ;) – 2017-01-02
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    Can you please clarify this part for me? $\left | \left | z \right |^{2k}-\left | 1 \right | \right |=1-\left | z \right |^{2k}$ – 2017-01-03
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    *By definition*, $|a|=a$ if $a \ge 0$ and $|a|=-a$ if $a\le 0$. Here $\left| \left| z \right|^{2k}-\left| 1 \right| \right|=-( \left| z \right|^{2k}-\left| 1 \right| )=1-\left| z \right|^{2k}$ since $\left| z \right|^{2k}-\left| 1 \right| \le 0$. Right? – 2017-01-03
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    Ah I see. Thank you so much! =) – 2017-01-03