Tauber theorem? 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 $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {{a_k}}  = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)f(x)$.
 A: 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 $(1-x)f(x)\to0$.
A: $\sum_{k=0}^{n}a_k\to\sum_{k=0}^{\infty}a_k$ as $n\to\infty$
$\frac{\sum_{k=0}^{\infty}a_k}{f(x)}\to 1$ as $x\to 1$
$(n+1)(1-x)=1-x+n-nx=0$ when $x=1$
$\lim {(n+1)(1-x)}=\lim {\frac{\sum_{k=0}^{n}a_k}{f(x)}}$
multiply both sides by $\frac{f(x)}{n+1}$
lim $\frac{1}{n+1}\sum_{k=0}^{n}a_k= \lim(1-x)f(x)$
A: the proof of this fact is presented in details in Tauberian Theory: a century of developments by Jacob Korevaar. In this text the result you are looking for is the Theorem 12.1 at page 22. 
Note that in your case the second Tauberian condition of the Theorem is satisfied, with $A$ (there) possibly being infinity.
 Edition  As observed by the questioner the theorem quoted above, give us just a partial answer for this question.
