Let $\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ be a power series. Show sum-function $g(z)$ is continuous on $|z|\le 1$. Let $\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ be a power series.
I've shown that radius of convergence is $R=1$.
I've a theorem saying that the sum-function $g(z)=\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ is continuous for $|z| < R$. 
How can I show that $g(z)$ is continuous on the set $|z|\le 1$ ?

This is only a partial answer taking into account comments and answers to comments.
I suppose that you noticed that $$f(z)=\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$$ is the antiderivative of $$\sum^{\infty}_{n=1} \frac {z^{n}} {n}=-\log (1-z)$$ So, integration by parts leads to $$f(z)=z+(1-z) \log (1-z)$$
