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How can we prove that the limit of sequence $\lim_{n\rightarrow\infty}a_{n}=0$ if $a_{n}>0$ and $\lim_{n\to\infty}\frac{a_{n}}{a_{n+1}}=l>1$

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What have you tried? –  Alex Becker May 24 '12 at 13:36
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3 Answers

up vote 3 down vote accepted

There is some $N$ sufficiently large so that, for all $n > N$, we have $a_{n+1}/a_n \leq c < 1$.

For for $n > 0$ we have $0 < a_{n+N} \leq a_N c^n$. By the squeeze theorem, $a_{n+N} \rightarrow 0$.

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The key observation in applying the squeeze theorem is $a_{N}$ is fixed. –  toypajme May 24 '12 at 14:18
    
I don't certainly get the point that how we reach the inequality $a_{n+N}<a_{N}c^n$ thanks –  Francis King May 24 '12 at 15:09
    
I have get the point.thanks –  Francis King May 24 '12 at 15:19
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Hint:

If $\lim\limits_{n\rightarrow\infty}{a_n\over a_{n+1}}=l>1$, then $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n }={1\over l}<1$. Choose $0<c<1$, and choose $N$ so that if $n\ge N$, then ${a_{n+1}\over a_n}<c$.

Then note:

$\ \ \ \ a_{N+1}<{c}\, a_N$,

$\ \ \ \ a_{N+2}<{c}\,a_{N+1}<{c^2}a_N$,

$\ \ \ \ a_{N+3}<{c}\,a_{N+2}<{c^3}a_N$,

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$

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thanks. your step-by-step induction helps me a lot. –  Francis King May 24 '12 at 15:20
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Another proof not-by-the-definition: the series $\,\displaystyle{\sum_{n=1}^\infty a_n\,}$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $\,\displaystyle{\frac{a_{n+1}}{a_n}\to \frac{1}{l}<1}\,$ , thus it must be $\,a_n\to 0$

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Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning. –  Siminore May 24 '12 at 16:27
    
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges. –  DonAntonio May 24 '12 at 18:01
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