The power series $\sum\limits_{n=1}^{\infty} \frac {z^{n} }{ n^{2}} \ $ This is an exercise from Remmert.
The power series $\sum\limits_{n=1}^{\infty} \frac {z^{n} }{ n^{2}} \ $ has radius of convergence $1 \ $. Show that the function it represents is injective in $\{ z \in \mathbb{C} | \ \ \lVert z \rVert < \frac{2}{3}  \} \ $.
The text gives the hint: $z^n -w^n = (z-w)\ ( z^{n-1}+z^{n-2}w + \ldots + w^{n-1} ) \ $.

As @TMM has written, start with the hint. If $z\neq w$, then
$$
\sum_{n=1}^\infty \frac{z^{n-1}+\cdots + w^{n-1}}{n^2} =0 \enspace. \quad (\star)
$$
The first term (with $n=1$) in the sum is actually $1$. The other terms are bounded, whenever $|z|, |w| \leq 2/3$, by
$$
\sum_{n\geq 2} \frac{n(2/3)^{n-1}}{n^2} = \frac32 \sum_{n\geq 2} \frac{(2/3)^{n}}n
= \frac32 \left(-\log\left(1-\frac23\right) - \frac23\right) =\frac32 \log3 -1 <1 \enspace,
$$
therefore $(\star)$ is never true if $|z|, |w| \leq 2/3$.
