Prove or provide a counter example: If $a_n \rightarrow 0$ and $\dfrac {a_{n+1}} {a_n} \rightarrow L$ then $L \in [-1;1]$. I think this is right and try to use contradiction to prove it but I still have no idea about this.
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For $n \ge N,\frac{a_{n+1}}{a_n} \in (L-\epsilon,L+\epsilon) \implies \frac{a_{N+k}}{a_N} \in \ldots$ (think to a geometric progression) – reuns Sep 21 '17 at 04:26
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Sorry, but I dont understand your answer. – Dota2 Sep 21 '17 at 04:28
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For every $\epsilon> 0$ there is a $N$ such that $ \frac{a_{N+k}}{a_N} \in ( (L-\epsilon)^k,(L+\epsilon)^k)$ – reuns Sep 21 '17 at 04:37
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Think about divergent series. – Nosrati Sep 21 '17 at 04:38
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The question that is asked is : Assume that $a_n \to 0$ and $\frac{a_{n+1}}{a_n}$ converges to $L$, and we are to show that $-1 \leq L \leq 1$.
Indeed, suppose that $|L| > 1$. Then, since $\lim \frac{a_{n+1}}{a_{n}} = L$, there exists $M$ large enough so that $\left|\frac{a_{N+1}}{a_N}\right| > 1$ for all $N > M$. This implies that for all $N > M$, we have $|a_{N+1}| > |a_N|$. Clearly, the fact that $\frac{a_{n+1}}{a_n}$ converges to $L$ means that it is bounded, shows that $a_n$ is non-zero, therefore $|a_M| > 0$, hence for all $N > M$, we have $|a_N| > |a_M|$. This contradicts the fact that $a_n \to 0$, since after a certain point no point of the sequence is closer to zero than $|a_M|$.
Hence, $|L| \leq 1$ must happen.
Sarvesh Ravichandran Iyer
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That's a real relief. I thought I interpreted the question incorrectly. Thank you. – Sarvesh Ravichandran Iyer Sep 21 '17 at 05:11