Abel summability under the hypothesis $a_n \ge 0$ Let $f(x) = \sum a_nx^n$, with $a_n \ge 0$, have radius of convergence $1$.  Show that if $\lim_{x \to 1^-} f(x) = \sigma$, then $\sum a_n$ converges and $\sigma = \sum a_n$.  This is Abel summability with the hypothesis that $a_n \ge 0$.
I'm not sure how to proceed.
 A: We don't need Abel's theorem here, since we have an ebarrassment of riches. For any $N$ we have
$$ \sum_{n=1}^{N}a_nx^n \le \sum_{n=1}^{\infty}a_nx^n $$
for $0< x < 1.$ Taking limits as $x \to 1^-$ above gives $\sum_{n=1}^{N}a_n \le \sigma.$ Since $N$ is arbitrary, we have 
$$\tag 1 \sum_{n=1}^{\infty}a_n \le \sigma.$$
On the other hand we clearly have, for $0<x<1,$
$$\sum_{n=1}^{\infty}a_nx^n \le \sum_{n=1}^{\infty}a_n.$$
Taking the limit as $x\to 1^-$ then shows
$$\tag 2 \sigma \le \sum_{n=1}^{\infty}a_n.$$
The inequalities $(1)$ and $(2)$ give the result.
A: Presumably you know the converse: If $\sum a_n=a$, with no restrictions on $a_n$, then $\lim_{x\to1}f(x)=a$. That's Abel's theorem.
Results saying $\lim f(x)=a$, with some restrictions on $a_n$, imply $\sum a_n=1$ are "tauberian" theorems. This one, assuming $a_n\ge0$, is a fairly simple one. Because $a_n\ge0$ implies that $\sum a_n$ exists; now if $\sum a_n=b$ then Abel's Theorem shows that $\lim f(x)=b$.
(To make that completely correct we need Abel's Theorem in the case $a_n\ge0$ and $\sum a_n=\infty$. But that follows by very much the same proof as the version for $\sum a_n=a\in\Bbb R$.)
