Let $(A_n)$ be a sequence of positive real number such that $$ \lim_{n\to\infty}\frac{A_{n+1}}{A_n}=r<1, $$ can we evaluate the $\lim\limits_{n\to\infty}A_n$?
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2Hint: this implies that the series $\sum A_n$ converges. What does this tell you about $\lim_{n \to \infty}A_n$? – Jul 14 '16 at 16:57
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yes n tends to infinite – Manzil Pradhan Jul 14 '16 at 17:30
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Possible duplicate of Showing that $a_n \to 0$ if $a_n/a_{n+1} \to l > 1$. – Jendrik Stelzner Sep 07 '18 at 10:46
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Supposing$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = r < 1$, we can select an $m$ such that $r < m < 1$. The limit implies that there must exist an $s \in \mathbb{N}$ such that we have $\displaystyle \frac{a_{n+1}}{a_n} < m$ for all $n \geq s$. In particular, $a_{s+1} < ma_s$.
Taking this a step further and applying the above, we also have $a_{s+2} < m a_{s+1} < m^2a_s$. In general, $a_{s+k} < m^ka_s$.
What does this tell us about $a_n$ as $n \rightarrow \infty$?
Note: This is essentially the beginning of the proof of the ratio test for the convergence of $\displaystyle \sum a_n$, where $a_n$ is a sequence as above.
Kaj Hansen
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You have twice been given information about the relationship of what you are given to convergence of the series (i.e. sum) and asked "what does that tell you about the limit of the sequence- and you have twice responded "yes, n tends to infinity". Yes, everyone knew that from the start! The question is "if $\sum A_n$ converges, what must be true of the sequence $A_n$?" – user247327 Jul 15 '16 at 15:27