I have the following homework:
Let $P(x) = \sum_{k = 0}^{\infty} a_k x^k$ be a power series with finite radius of convergence $r > 0$ such that $P(r)$ diverges and $a_k \geq 0$ for all $k \in \mathbb{N}$. Show that:
$$\lim_{x \rightarrow r-} P(x) = \infty$$
My attempt at a solution:
We have shown Abel's theorem in class. So I know that $P(x)$ is continuous at $x = r$. Shouldn't it follow immediately then that $\lim_{x \rightarrow r-} P(x) = P(r) = \infty$, because $P(r)$ diverges ?
Because there are $5$ marks for this question it certainly won't work this way. What am I overlooking?
For any $M>0$, there is $N>0$ such that $$ \sum_{k\leq N} a_kr^k \ge M. $$
Consider $$ \lim_{x\rightarrow r-}\sum_{k\le N} a_k x^k=\sum_{k\leq N} a_k r^k \geq M. $$
Since $a_k\geq 0$, for any $0\leq x <r$,
$$ \sum_{k=0}^{\infty} a_k x^k \geq \sum_{k\leq N} a_k x^k $$ Now, we take $\liminf_{x\rightarrow r-}$ both sides, then $$ \liminf_{x\rightarrow r-}\sum_{k=0}^{\infty} a_k x^k \geq \liminf_{x\rightarrow r-}\sum_{k\leq N} a_k x^k = \lim_{x\rightarrow r-}\sum_{k\leq N} a_k x^k \geq M. $$ Therefore, $$ \lim_{x\rightarrow r-} P(x)=\infty. $$