TITLE: Tauber theorem?
QUESTION [6 upvotes]: Let $f(x) = \sum\limits_{k = 0}^\infty  {{a_k}{x^k}} $, where $a_k\ge0 $, and when $0 \le x < 1$ , $f(x)$ converges. Then please show that $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {{a_k}}  = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)f(x)$.

REPLY [4 votes]: The hypothesis is that $A_n=(n+1)C+o(n)$ for a given $C$, with $A_n=\sum\limits_{k=0}^na_k$. Thus, for every positive $\varepsilon$ there exists some finite $B_a$ and $B'_a$ such that $(C-\varepsilon)(n+1)+B_a\leqslant A_n\leqslant (C+\varepsilon)(n+1)+B'_a$ for every $n$. To use this fact, rewrite $f(x)$ for every $x$ in $[0,1)$ as
$$
f(x)=\sum\limits_{n=0}^{+\infty}A_nx^n(1-x).
$$
Since $\sum\limits_{n=0}^{+\infty}x^n(1-x)=1$ and $\sum\limits_{n=0}^{+\infty}(n+1)x^n(1-x)=\frac1{1-x}$, this yields, for every $x$ in $[0,1)$,
$$
C-\varepsilon+B_a(1-x)\leqslant(1-x)f(x)\leqslant C+\varepsilon+B'_a(1-x).
$$
Since $a$ is as small as one wants, this proves that $(1-x)f(x)\to C$ when $x\to1$, $x<1$.
Notes: (1) The hypothesis that $a_n$ is nonnegative is not needed. (2) If the limit $C$ of $\frac1{n+1}A_n$ is $-\infty$ or $+\infty$, one-sided versions of the proof above still apply to show that $(1-x)f(x)\to C$ when $x\to1$, $x<1$.
Edit (This is to answer a question asked by the OP in a comment.)
If $a_n$ may be positive or negative, $(1-x)f(x)$ may have a finite limit when $x\to1$, $x<1$, although $C_n=\frac1{n+1}A_n$ diverges. For example, if $a_n=(-1)^{n+1}n$ for every nonnegative $n$, the limit points of $(C_n)_n$ are $\pm\frac12$ hence $(C_n)_n$ diverges but $f(x)=\frac{x}{(1+x)^2}$ hence $(1-x)f(x)\to0$.