The following classical result is due to E. Cesàro:
Let $\left\{ a_{n}\right\} _{n\geq0}$ be a sequence of complex numbers and let $\left\{ b_{n}\right\} _{n\geq0}$ be a sequence of positive real numbers. Suppose that the power series $\sum_{n=0}^{\infty}a_{n}z^{n}$ and $\sum_{n=0}^{\infty}b_{n}z^{n}$ both converge for $\left|z\right|<1$, and that the $b_{n}$ series diverges for $z=1$. Then:
$$\lim_{r\uparrow1}\frac{\sum_{n=0}^{\infty}a_{n}r^{n}}{\sum_{n=0}^{\infty}b_{n}r^{n}}=\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$$ whenever $\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ exists.
I ask: does the converse hold?
That is, assuming: $$\lim_{r\uparrow1}\frac{\sum_{n=0}^{\infty}a_{n}r^{n}}{\sum_{n=0}^{\infty}b_{n}r^{n}}$$ exists and equals $c$, does it then follow that $\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ exists, and equals $c$?
It's certainly possible for the limit not to exist, see for instance $a_n=\begin{cases}1&\text{if }2\mid n\\ 2^{-n}&\text{if } 2\nmid n\end{cases}$ and $b_n=1$. Then \begin{align}&\sum_{n=0}^\infty a_nr^n=\frac1{1-r^2}+\frac{2r}{4-r^2}\\&\sum_{n=0}^\infty b_nr^n=\frac1{1-r}\\ &\lim_{r\uparrow 1}\frac{\sum_{n=0}^\infty a_nr^n}{\sum_{n=0}^\infty b_nr^n}=\frac12\\ &\limsup_{n\to\infty}\frac{a_n}{b_n}=1\\&\liminf_{n\to\infty}\frac{a_n}{b_n}=0\end{align}