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If $d(x_{n+1},x_n)<\frac{1}{3^n}$, then, is $x_n$ convergent?

3 Answers 3

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Consider $m > n$. Then by triangle inequality $d(x_m,x_n) \leq d(x_m,x_{m-1}) + d(x_{m-1},x_{m-2}) + \cdots + d(x_{n+1},x_n) \\ = \dfrac1{3^{m-1}} + \dfrac1{3^{m-2}} + \cdots + \dfrac1{3^n} < \sum_{k=n}^{\infty} \dfrac1{3^k} = \dfrac1{3^n} \dfrac32 = \dfrac1{2 \cdot 3^{n-1}}$ Use this to prove that the sequence is a Cauchy sequence, and make use of completeness of $\mathbb{R}$ to prove that the sequence converges.

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    @jack Yes. Thats true. Given \epsilon > 0, choosing $N(\epsilon) = -\log_3(2 \epsilon) + 1$, we have that for all m,n > N(\epsilon), \vert x_m - x_n \vert < \epsilon. Hence, Cauchy and hence converges.2012-10-15
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HINTS: Suppose that $|x_{n+1}-x_n|<3^{-n}$ for all $n\in\Bbb N$. Let $m\in\Bbb N$ be arbitrary; what can you say about $\sum_{n\ge m}\frac1{3^n}\:?$ What does this imply about $|x_k-x_n|$ for $k,n\ge m$? Finally, what do you know about Cauchy sequences in $\Bbb R$?

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    Thanks for your help as well.2012-10-15
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You can use the fact that $\sum_{n=0}^{\infty} |a_{n+1}-a_{n}|< \infty $ to prove the sequence is Cauchy and hence convergent in $\mathbb{R}$. See here.