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October 9th, 2008, 10:04 PM  #1 
Member Joined: Sep 2008 Posts: 46 Thanks: 0  cauchy sequence that isn't a fast cauchy sequence
in R with the usual definition of distance, let x_n be a fast cauchy sequence if sum_n_inf d(x_n+1,x_n) < infinity i'm looking for {x_n} such that it is cauchy but isn't fast. my initial guesses were x_n = 1/n and x_n = sqrt(n+1) sqrt(n) for the latter guess, I need to show that sum of sqrt(n+1)2sqrt(n)+sqrt(n1) blows up. any thoughts? also, i need to find a similar sequence X_n in Rn such that X_n converges but isn't fast. Thanks. 
October 10th, 2008, 02:46 AM  #2  
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7  Re: cauchy sequence that isn't a fast cauchy sequence Quote:
sqrt(n+1)*(12sqrt(11/(n+1))+sqrt(12/(n+1))) = sqrt(n+1)*[12*(11/(2(n+1))1/(8(n+1)^2)+o(1/(n+1)^2))+11/(n+1)1/(2(n+1)^2)+o(1/(n+1)^2)] = sqrt(n+1)*(1/(4(n+1)^2)+o(1/(n+1)^2)) Therefore sqrt(n+1)2sqrt(n)+sqrt(n1) ~ 1/(4*(n+1)^(3/2)), which unfortunately is the general term of a convergent series; this means that your initial statement doesn't hold for this instance. Your first example holds, though.  
October 10th, 2008, 05:28 AM  #3 
Member Joined: Sep 2008 Posts: 46 Thanks: 0  Re: cauchy sequence that isn't a fast cauchy sequence
You replied that the x_j = 1/j allows for cauchy but not fast cauchy... how do we know this? it is obvious that it is cauchy, but to show it is fast cauchy: d(x_j+1,x_j) = 1/(j+1)  1/j = 1/(j^2+j) the sum of 1/(j^2+2j) <= sum of 1/j^2 which converges... so isn't this fast cauchy as well? not sure how to get my cauchy, but not fast. how would i also come up with such a sequence of points {X_j} in Rn that converge but don't satisfy fast property? 
October 10th, 2008, 07:49 AM  #4 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7  Re: cauchy sequence that isn't a fast cauchy sequence
Yeah, that's actually what I should have meant, (1/n)_n is also fast Cauchy (the sum of the distances between two consecutive terms is a telescopic sum, therefore you don't even need to find an upper bound as you did in order to show the convergence). To answer your initial question, you can consider the following sequence which is Cauchy but not fast Cauchy: u_n = (1(1)^n)*1/n 
October 10th, 2008, 08:27 AM  #5 
Member Joined: Sep 2008 Posts: 46 Thanks: 0  Re: cauchy sequence that isn't a fast cauchy sequence
Thanks. I see how your example is Cauchy, but I'm trying to verify it is fast cauchy. I must evaluate the sum of 2^(1/n), correct? We see that 2^1 + 2^1/2 +..., and thus we can see that 2^1/3+2^1/4 > 2*2^1/4 > 2.... and thus we have a summation of 2's? So for Rn, we just define a component as this sequence and rest 0, thus the distance between any two points is >= to the distance for the specific component, which does not converge so the points don't converge... is this correct? 
October 10th, 2008, 08:41 AM  #6 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7  Re: cauchy sequence that isn't a fast cauchy sequence
Sorry, I meant (1(1)^n)/(2n), so that you end up with the sequence 0,1,0,1/4,0,1/6,0, etc .... we could take the sequence 0,1,0,1/2,0,1/3,0,1/4,0,1/5, etc ... as well.

October 10th, 2008, 12:11 PM  #7 
Member Joined: Sep 2008 Posts: 46 Thanks: 0  Re: cauchy sequence that isn't a fast cauchy sequence
Yeah. Thanks. I wrote it up with your first suggestion and proved that it wasn't fast cauchy as the nonzero terms are always greater than 1... then I thought, #$%#!!! then it isn't even cauchy. Your next suggestion works but it wasn't obvious to me that the harmonic series for even n is divergent. But I guess the easiest way to come up with a sequence is just to outline how the terms look like instead of what the general formula is. So 0,1,0,1/2,0,1/3,0,... is technically equivalent to the harmonic series then right? Thanks! 

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