Intuition behind stolz caesaro theorem The stolz caesaro theorem seems to be a discrete analogue of L'hopitals rule. I can understand l'hopitals rule via the taylor series, that is:
$$ \lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{ P(a) + \frac{dP}{dx}|_a (x-a) + O ( (x-a)^2)  }{  Q(a) + \frac{dQ}{dx}|_a (x-a) + O ( (x-a)^2)}$$
Now if $P(a) = Q(a) = 0$, the limit becomes ratio of first order term/ the ratio of values nearby to the point of interest. So, Is there a similar 'series' intuition for stolz caesaro theorem?
 A: It seems trivial compared to Taylor series or L'Hopital, because it follows directly from a much more general theorem:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Theorem. Take any $a,b,c∈ℂ$ and sequences $f,g$ from $ℂ$ such that both $\sum_{k=0}^n g(k) → ∞$ and $\lfrac{f(n)}{g(n)} → c$ as $n→∞$. Then $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) } → c$ as $n→∞$.
Proof. For convenience let $[ε]$ denote $\{ z : z∈ℂ ∧ |z|<ε \}$. Take any $ε>0$. Let $m∈ℕ$ such that $\lfrac{f(n)}{g(n)} ∈ c+[ε]$ for every $n∈ℕ_{≥m}$. As $n→∞$, $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) }$ $= \lfrac{ a + \sum_{k=0}^{m-1} f(k) + \sum_{k=m}^n f(k) }{ b + \sum_{k=0}^{m-1} g(k) + \sum_{k=m}^n g(k) }$ $\overset{*}∈ \lfrac{ (1+o(1)) · \sum_{k=m}^n f(k) }{ (1+o(1)) · \sum_{k=m}^n g(k) }$ $⊆ (1+o(1)) · \lfrac{ \sum_{k=m}^n \big( (c+[ε])·g(k) \big) }{ \sum_{k=m}^n g(k) }$ $⊆ (1+o(1)) · (c+[ε])$, where the $*$-marked relation holds because $\sum_{k=m}^n f(k) → ∞$ and $\sum_{k=m}^n g(k) → ∞$. Since this holds for every $ε>0$, therefore $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) } → c$ as $n→∞$.
−−−
Note that in the above theorem statement we allow division by zero but the assertion of a limit means that eventually the value is defined (no division by zero). If you wish to avoid that issue, simply impose that $g(n)$ and $b + \sum_{k=0}^n g(k)$ are never zero for every $n∈ℕ$.
