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How to prove the $0/0$ case of Stolz-Cesaro Theorem? In other words:

Given that $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = 0$$ with $(b_n)_{n=1}^{\infty}$ strictly monotone, and that $$\lim_{n \to \infty} \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = L$$ prove that $\lim_{n \to \infty} \frac{a_n}{b_n} = L$.

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  • $\begingroup$ You find here a proof for that theorem. $\endgroup$ – Salech Rubenstein Dec 9 '13 at 4:01
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    $\begingroup$ But the proof is for the case when $\lim_{n \to \infty} b_n = +\infty$. $\endgroup$ – AaronS Dec 9 '13 at 4:39
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Without loss of generality, suppose ${b_n}$ monotonically decreases. Thus $b_n\gt0$ and $b_n-b_{n+1} \gt 0$.
If $$\lim_{n\to \infty}\frac{a_n-a_{n+1}}{b_n-b_{n+1}}=L,$$ then for any given $\epsilon \gt 0$, there exists a natural number $N$ such that $$(L-\epsilon)(b_n-b_{n+1}) \lt a_n-a_{n+1} \lt (L+\epsilon)(b_n-b_{n+1})$$ for all $n \gt N$.
Let $k \gt n$ be a natural number. Summing the inequalities above, we get: $$(L-\epsilon)\sum_{i=n}^{k}(b_i-b_{i+1}) \lt \sum_{i=n}^{k}(a_i-a_{i+1}) \lt (L+\epsilon)\sum_{i=n}^{k}(b_i-b_{i+1}),$$ i.e.,$$(L-\epsilon)(b_n-b_{k+1}) \lt a_n-a_{k+1} \lt (L+\epsilon)(b_n-b_{k+1}).$$ In this inequality, fix $n$ and let $k \to \infty$, we get $$(L-\epsilon)b_n \le a_n \le (L+\epsilon)b_n,$$ that is, $$L-\epsilon \le \frac{a_n}{b_n} \le L+\epsilon.$$ This relation holds for all $n \gt N$, so finally $$\lim_{n \to \infty}\frac{a_n}{b_n}=L.$$

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  • $\begingroup$ why u can switch the order of $\frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ to $\frac{a_{n} - a_{n+1}}{b_{n} - b_{n+1}}$ $\endgroup$ – user10024395 Apr 21 '14 at 8:44
  • $\begingroup$ why u can assume bn monotone decreasing? $\endgroup$ – user10024395 Apr 21 '14 at 8:47
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    $\begingroup$ @user136266 If $(b_n)$ is increasing, let $c_n=-b_n$, then $(c_n)$ is decreasing. Apply the proof to $(c_n)$, you'll get the same conclusion for $(b_n)$. $\endgroup$ – AaronS Apr 21 '14 at 10:08

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