How can we prove that $\lim_{n\rightarrow\infty}a_{n}=0$ if $a_n>0$ and $\lim_{n\to\infty} a_n / a_{n+1} = l >1$?
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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, $a_{n+N} \rightarrow 0$.
David Harris
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The key observation in applying the squeeze theorem is $a_{N}$ is fixed. – toypajme May 24 '12 at 14:18
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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
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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:
$\ \ \ \ a_{N+1}<{c}\, a_N$,
$\ \ \ \ a_{N+2}<{c}\,a_{N+1}<{c^2}a_N$,
$\ \ \ \ a_{N+3}<{c}\,a_{N+2}<{c^3}a_N$,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
David Mitra
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in first step it s $\frac{x_{n+1}}{x_n} < l + \epsilon$. how have you taken c here. if we take $c = l +\epsilon$, then for c <1 we need to have $\epsilon < 1-l$. is it right ? – Olivia Apr 06 '23 at 06:41
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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 $\,a_n\to 0$
DonAntonio
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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
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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. – DonAntonio May 24 '12 at 18:01