My Math Forum Cauchy Sequence II

 Complex Analysis Complex Analysis Math Forum

November 10th, 2016, 02:49 PM   #1
Member

Joined: Nov 2016
From: Kansas

Posts: 73
Thanks: 1

Cauchy Sequence II

How to solve this?
Attached Images
 Capture7.PNG (9.4 KB, 9 views)

 November 10th, 2016, 04:42 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 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 04:50 PM.
November 11th, 2016, 02:06 PM   #3
Global Moderator

Joined: May 2007

Posts: 6,510
Thanks: 584

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.

 Tags cauchy, sequence

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ZMD Real Analysis 1 November 10th, 2016 02:17 PM xdeimos Real Analysis 1 October 11th, 2013 10:29 PM jrklx250s Real Analysis 5 December 12th, 2011 06:20 PM rose3 Real Analysis 1 November 3rd, 2009 01:26 PM babyRudin Real Analysis 6 October 10th, 2008 11:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top