I'm having some difficulty with this question:
Prove that the function $\,\,\displaystyle f(z) = \sum_{n=1}^\infty\frac{z^n}{n^2}\,$ is univalent in the disk $D\left(\dfrac23\right)$.
There is the following hint: $\dfrac{z^n-w^n}{z-w} = z^{n-1} + wz^{n-2} + ... + w^{n-2}z + w^{n-1}.$
Any ideas?
Assume that $f(z)=f(w)$, for $z\ne w$ and $\lvert z\rvert,\,\lvert w\rvert<\frac23$. Then $$ 0=\sum_{n=1}^\infty\frac{z^n-w^n}{n^2}=(z-w)\sum_{n=1}^\infty\frac{z^{n-1+}z^{n-2}w+\cdots+w^{n-1}}{n^2}. $$ Hence $$ 0=1+\sum_{n=2}^\infty\frac{z^{n-1+}z^{n-2}w+\cdots+w^{n-1}}{n^2}. $$ But $$ \sum_{n=2}^\infty\frac{\lvert z^{n-1+}z^{n-2}w+\cdots+w^{n-1}\rvert}{n^2}\le \sum_{n=2}^\infty\frac{n(2/3)^{n-1}}{n^2}=\sum_{n=2}^\infty\frac{(2/3)^{n-1}}{n}=-\frac{3}{2}\big(1+\ln (1-2/3)\big)<1. $$ Since, for $0<\lvert x\rvert<1$, $$ \sum_{n=2}^\infty\frac{x^{n-1}}{n}=\frac{1}{x}\left(\sum_{n=1}^\infty\frac{x^{n}}{n}-1\right)=\frac{1}{x}(-\ln(1-x)-1). $$ Note. This $f$ is univalent even in $D\big(\frac{5}{6}\big)$.