Exericise 9.36: If each $a_n >0$ and $\sum a_n$ diverges, prove that $\sum a_n x^n \to +\infty$ as $x\to1^-$. (Assuming $\sum a_n x^n$ converges for $|x|<1$.
Here's what I tried:
Since $\sum a_n$ diverges, for $M>0$, $\exists N$ such that $\sum_{k=0}^N a_k >M.$ I want to show that for $M>0$, there exists $\delta$ such that, whenever $\delta<x<1$ implies that there exists $p$ such that $\sum_{k=0}^p a_k>M$.
I was reading this solution and I don't get how the "then by continuity" part implies (**) part. Could someone explain? Thanks!
Let $f_n(x)=\sum_{k=1}^n a_nx^n$. Since this is continuous, $\forall\epsilon>0\,\exists\delta<1$ such that $\delta<x<1$ implies $$\sum_{k=1}^{n} a_k-\sum_{k=1}^{n} a_kx^k<\epsilon$$
Let $\epsilon=\sum a_k-M$. Choose $N$, such that $\sum_{k=1}^{N} a_k>M$. Now,
$$\sum_{k=1}^{N} a_kx^k>\sum_{k=1}^{N}a_k-\epsilon=M.$$