Prove that: if $x_n \gt 0\forall n\in \mathbb N$ and if $\lim_{n\rightarrow \infty}\frac{x_{n +1}}{x_n} \lt 1$ then $\lim_{n\to \infty}x_n =0$ I don't know how to start, please help me.
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Intuitively... $(x_n)$ is a decreasing sequence and bounded by $0$... does this say something about $(x_n)\rightarrow ??$ – Nov 05 '13 at 05:12
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1The same can be said about $;\frac{n+1}n;$, @PraphullaKoushik ...something else is lacking. – DonAntonio Nov 05 '13 at 05:13
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@DonAntonio : for $\frac{n+1}{n}$ we do not have $lim _{n\to \infty }\frac{n+1}{n}<1$ we actually have $lim _{n\to \infty }\frac{n+1}{n}=1$.. so, the example which you have given does not fit into given condition... – Nov 05 '13 at 06:12
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I know my example's limit isn't less than one, @PraphullaKoushik : I meant your statement about hte seq. being decreagin and bounded by zero... – DonAntonio Nov 05 '13 at 13:10
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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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Hint. Let $L=\lim x_{n+1}/x_n$. By definition of limit, there is $N$ such that $n>N$ implies $|x_{n+1}/x_n -L|<(1-L)/2$. So if $n>N$ then $x_{n+1}/x_n<(1+L)/2$.
Hanul Jeon
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1I can´t see how from this, I get that ${x_n}$ tends to 0 (zero) I am trying to figure it out, can you tell me another hint? – David Hernandez Nov 05 '13 at 05:38
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@DavidHernandez So if $n>N$ then $x_n<((1+L)/2)x_{n-1}<\cdots<((1+L)/2)^{n-N} x_N$. Note that $(1+L)/2<1$ and $N$ is fixed number. – Hanul Jeon Nov 05 '13 at 05:42
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W.l.o.g. by ignoring the first few terms we have $\frac{x_{n+1}}{x_n} \le a < 1$ for all $n$.
Then $x_{n+1} \le a^n x_1 \rightarrow 0$.
Deven Ware
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Also note, this method is how you prove the ratio test used in DonAntonio's answer. – Deven Ware Nov 05 '13 at 05:31
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The exercise is easily solved with the help of infinite series:
$$\text{The series}\;\;\sum_{n=1}^\infty x_n\;\;\text{is a positive one}$$
and we're given that the ratio test renders the series convergent, so it must be
$$\lim_{n\to\infty} x_n=0$$
DonAntonio
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