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November 10th, 2016, 03:49 PM   #1
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Cauchy Sequence II

How to solve this?
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 November 10th, 2016, 05:42 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,445 Thanks: 2117 Math Focus: Mainly analysis and algebra The obvious approach would be to derive the closed form for the sum $$s_n = \frac{1-x_0^{n+1}}{1-x_0}$$and then show that $s_n$ is Cauchy. I would begin by noting that for any given $N$, we have $$s_N \lt s_n \lt \frac1{1-x_0} \quad \text{for all n \gt N}$$ Thus, for any given $\epsilon$, we can pick $N$ such that $\frac1{1-x_0}-s_N \lt \epsilon$. Note that the above only works for $0 \le x_0 \lt 1$, but the approach can be easily modified for $-1 \lt x_0 \lt 0$. Last edited by v8archie; November 10th, 2016 at 05:50 PM.
November 11th, 2016, 03:06 PM   #3
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Quote:
 Originally Posted by v8archie The obvious approach would be to derive the closed form for the sum $$s_n = \frac{1-x_0^{n+1}}{1-x_0}$$and then show that $s_n$ is Cauchy. I would begin by noting that for any given $N$, we have $$s_N \lt s_n \lt \frac1{1-x_0} \quad \text{for all n \gt N}$$ Thus, for any given $\epsilon$, we can pick $N$ such that $\frac1{1-x_0}-s_N \lt \epsilon$. Note that the above only works for $0 \le x_0 \lt 1$, but the approach can be easily modified for $-1 \lt x_0 \lt 0$.
The ratio is not $\displaystyle x_0$ but $\displaystyle \frac{1}{x_0}$. I suspect something wrong in problem statement.

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