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I'm trying to understand a step from a proof of the Stolz-Cesaro Theorem.

Let ${\left\{ {{b_n}} \right\}_{n \in {\Bbb N}}}$ is a positively strictly increasing unbounded sequence.

If ${\left\{ {{a_n}} \right\}_{n \in {\Bbb N}}}$ is another sequence and

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} = l $$

then

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = l$$

The proof I'm trying to understand is here. I can't seem to understand the last step, i.e. why: $$(l-\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})+\frac{a_{N(\epsilon)}}{b_{k+1}} < \frac{a_{k+1}}{b_{k+1}}<(l+\epsilon)(1-\frac{b_{N(\epsilon)}}{b_{k+1}})+\frac{a_{N(\epsilon)}}{b_{k+1}} \implies\\ \implies l-\epsilon<\frac{a_{k+1}}{b_{k+1}} < l + \epsilon$$

How is that true and how can one write that more formally and detailed?

Thanks in advance!

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    $\begingroup$ If $a_{n} \leq b_{n} $ then we have $\limsup a_{n} \leq \liminf b_{n} $ and the final outcome should be $$l-\epsilon \leq\liminf \frac{a_{k}} {b_{k}} \leq \limsup \frac{a_{k}} {b_{k}} \leq l+\epsilon $$ and result follows since $\epsilon$ is arbitrary. $\endgroup$ – Paramanand Singh Sep 23 '17 at 13:52
  • $\begingroup$ See discussion here: math.stackexchange.com/questions/109069/… $\endgroup$ – Hans Lundmark Sep 23 '17 at 13:58

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