A Tauberian theorem for a quotient of power series, the limit on the boundary 
Take sequences $a_n, b_n \in \mathbb R_{>0}$ (or $\mathbb C$) such that the limit $$L = \lim_{N \to \infty}\frac{\sum_{n \leq N} a_n }{\sum_{n \leq N} b_n}$$
  exists and such that the power series $f(x) = \sum a_n x^n$, $g(x) = \sum b_n x^n$ converge in $(-1,1)$. Is there any theorem (possibly with additional conditions) that guarantees that 
  $$\lim_{x \to 1}\frac{f(x)}{g(x)} = L \quad?$$
  in particular, that the limit exists?

I will be happy if this holds with the limit taken in the interval $[0,1)$, not necessarily in the full disk $B(0,1)$.
I'm especially interested in the case where $\sum_{n \leq N} a_n$ grows fast, e.g. polynomially with $N$.
There is an article with a promising title:
B. I. Korenblyum [B. Korenblum], "A general Tauberian theorem for a quotient of functions", Dokl. Akad. Nauk SSSR, 88 1953, 745–748, Russian.
But I have no access to it. (Volume 88 is missing in the online archive.)
 A: For non-negative coefficients and $x \in [0,1)$, it is rather straightforward using the theory of general Dirichlet series. For $t \geqslant 0$ define
$$A(t) = \sum_{n \leqslant t} a_n$$
and $B(t)$ analogously. Write $x = e^{-s}$ with $s > 0$. Since
$$e^{-\lambda s} = s\int_{\lambda}^{\infty} e^{-ts}\,dt$$
we can write
$$f(e^{-s}) = \sum_{n = 0}^{\infty} a_n e^{-ns} = s\sum_{n = 0}^{\infty} a_n \int_{n}^{\infty} e^{-ts}\,dt = s\int_0^{\infty} A(t)e^{-ts}\,dt$$
and the analogous expression for $g$.
If $L_1 < L < L_2$, then by assumption we have
$$L_1 B(t) \leqslant A(t) \leqslant L_2 B(t)$$
for $t \geqslant t_0$ and hence
$$L_1\int_{t_0}^{\infty} B(t) e^{-ts}\,dt \leqslant \int_{t_0}^{\infty} A(t) e^{-ts}\,dt \leqslant L_2 \int_{t_0}^{\infty} B(t) e^{ts}\,dt\,.$$
Thus
$$L_1\Biggl(s^{-s}g(e^{-s}) - \int_0^{t_0} B(t) e^{-ts}\,dt\Biggr) \leqslant s^{-1}f(e^{-s}) - \int_0^{t_0} A(t) e^{-ts}\,dt \leqslant L_2\Biggl(s^{-s}g(e^{-s}) - \int_0^{t_0} B(t) e^{-ts}\,dt\Biggr)$$
and upon division by $s^{-1}g(e^{-s})$ we obtain
$$L_1 + s\frac{L_1M(s)-K(s)}{g(e^{-s})} \leqslant \frac{f(e^{-s})}{g(e^{-s})} \leqslant L_2 + s \frac{L_2 M(s) - K(s)}{g(e^{-s})}$$
where
$$M(s) = \int_0^{t_0} B(t)e^{-ts}\,dt \qquad \text{and}\qquad K(s) = \int_{0}^{t_0} A(t) e^{-ts}\,dt\,.$$
Since $M$ and $K$ are entire functions, and $g(e^{-s})$ is monotonically decreasing with $\lim_{s \to 0^+} g(e^{-s}) \in (0, + \infty]$, the left and right hand sides of the above inequality tend to $L_1$ and $L_2$ respectively, whence
$$L_1 \leqslant \liminf_{s \to 0^+} \frac{f(e^{-s})}{g(e^{-s})} \leqslant \limsup_{s \to 0^+} \frac{f(e^{-s})}{g(e^{-s})} \leqslant L_2\,.$$
Since this holds for arbitrary $L_1 < L$ and $L_2 > L$, the existence of the limit follows.
The argument can easily be generalised to real $a_n, b_n$ under the condition that there is a $\beta > 0$ such that $B(t) \geqslant \beta$ for all sufficiently large $t$.
