Divergence of power series at the edge Let $\sum_{n=0}^\infty a_n x^n$ be a power series with radius of convergence $R \in (0,\infty)$ and $a_n \ge 0$ for all $n$.
Show that if $\sum_{n=0}^\infty a_n R^n = \infty$ then 
$\lim_{x \nearrow R} \sum_{n=0}^\infty a_n x^n = \infty$.
The theorem sounds intuitive but I don't know how to approach a proof. Can you give me a hint?

Take $M>0$. Since $\sum_{n=0}^\infty a_nR^n=+\infty$, there is a $N\in\mathbb N$ such that $\sum_{n=0}^Na_nR^n>M$. Therefore$$\tag1\lim_{x\to R^-}\sum_{n=0}^Na_nx^n=\sum_{n=0}^Na_nR^n>M.$$Since each $a_n$ is a non-negative real,$$\tag2\bigl(\forall x\in[0,R)\bigr):\sum_{n=0}^\infty a_nx^n\geqslant\sum_{n=0}^Na_nx^n.$$It follows from $(1)$ and $(2)$ that$$\lim_{x\to R^-}\sum_{n=0}^\infty a_nx^n=+\infty.$$
