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**hatsoff** Recall that the radius of convergence is the same for $\displaystyle |a_n z^n|$ as it is for $\displaystyle a_nz^n$. Then that $\displaystyle R\geq 1$ follows immediately from the hypotheses. Now suppose towards a contradiction that $\displaystyle R>1$, and choose $\displaystyle m\in(1,R)$. Then $\displaystyle \sum |a_n m^n|$ converges.

Notice that $\displaystyle \lim_{n\to\infty}\frac{(\ln n)/\ln m}{n}=\lim_{n\to\infty}\frac{1}{n\ln m}=0$, so we can choose $\displaystyle N$ such that $\displaystyle n>\ln n/\ln m$ and therefore $\displaystyle m^n>n$ for all $\displaystyle n\geq N$

It follows that $\displaystyle \sum_{n=N}^\infty |a_n m^n|\geq\sum_{n=N}^\infty n|a_n|=\infty$, a contradiction.