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How can we prove that $\lim_{n\rightarrow\infty}a_{n}=0$ if $a_n>0$ and $\lim_{n\to\infty} a_n / a_{n+1} = l >1$?

3 Answers 3

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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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    I have get the point.thanks2012-05-24
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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, and choose $N$ so that if $n\ge N$, then ${a_{n+1}\over a_n}.

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.2012-05-24
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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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    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.2012-05-24