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Cauchy sequence problem

  1. Sep 23, 2007 #1


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    Hey all, I've lurked on here and have found you all very useful and I have this question that is really bugging me.

    How would you prove something is a cauchy sequence using tits definition.
  2. jcsd
  3. Sep 23, 2007 #2


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    I am sorely tempted to say something about your typo but I will behave!

    You prove ANYTHING "using its definition" by using the precise formulas and words in the definition.

    The definition of "Cauchy sequence" is "{an} is a Cauchy sequence if and only if |an- am| goes to 0 as m and n go (independently) to infinity" or, in precise terms of the definition of limit, "{an} is a Cauchy sequence if and only if, given [itex]\epsilon > 0[/itex], there exist a positive integer N such that if m and n are both greater than N then |an- am< [itex]\epsilon[/itex]."

    The "independently" part is in the fact that m and n can be any integers larger than N- you can't just prove it for, say, consecutive terms an and an+1.

    If, for example, I wanted to prove that {1/n} is a Cauchy sequence, I would look at |1/n- 1/m|= |(m-n)/mn|. I would have to prove that that sequence converges to 0 as m and n go to infinity. I might do that by noting that |m-n|< m so |(m-n)/mn|< |m/mn|= |1/n|. To guarentee that that was "[itex]< \epsilon[/itex]" I would only have to ensure that N was larger than [itex]1/\epsilon[/itex].

    Of course, in that case, we are really proving that {1/n} goes to 0. It would be easier to prove that the original sequence converges and then use the fact that all convergent sequences are Cauchy sequences rather than use the definition.

    The importance of Cauchy sequence is that, in the real numbers, every Cauchy sequence converges. Since we know that, to prove a sequence is a Cauchy sequence, we must prove something goes to 0, that can be used to prove a sequence converges even if we don't know what it converges to.
  4. Sep 24, 2007 #3
    If any sequence in (X,d) is convergent, it is cauchy.

    Use convergence of x(n) smaller than epsilon/2 and use triangle inequality. This proof still requires some precision, so be careful.

    The idea of a Cauchy sequence is important is because it helps characterize completeness. If a sequence being cauchy implies it being convergent, the metric space is complete.
  5. Oct 22, 2010 #4
    If you deal with the real space R, then any Cauchy sequence converges. But, if not, then you could solve the problem by comparing the metric space with another metric space that is complete, e.g. R. (Recall that a metric space is called complete if and only if every Cauchy sequence converges).


    Prove that a metric space ((-pi/2, pi/2) , d) where d(x,y)=|tan x - tan y| is complete.

    Let x_n be a Cauchy sequence in ((-pi/2, pi/2) , d). Then, since tan is a continuous function in (-pi/2, pi/2), tan x_n is a Cauchy sequence in the image space. In this case the image is just the whole R. Since R is complete, it follows that y_n = tan x_n converges to a point in R, say y, i.e. |y_n - y| < [tex]\epsilon[/tex]. Let x = arctan y. Then
    d(x_n, x) = |tan x_n - tan x| = |y_n - y| < [tex]\epsilon[/tex]​
    and so ((-pi/2, pi/2) , d) is complete.
  6. Oct 23, 2010 #5


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    What a geezer... (-;
  7. Dec 12, 2010 #6
    How would you guys go about this one

    If 0<r<1 and |x(n+1) - x(n)| < r ^n for all n. Show that x(n) is Cauchy sequence.
  8. Dec 12, 2010 #7
    If someone gives you a number E > 0, can you find a corresponding n so that r^n < E ?
  9. Dec 13, 2011 #8
    we have |x_(n+1)-x_n|<r ^n for all natural n and 0<r<1

    we want te make |x_n-x_m|<ε by taking n and m sufficiently large. without loss of generality we can assume n<m. Then, defining k=m-n we have m=n+k so |x_n-x_m|=|x_n-x_(n+k)|

    Now, using the triangle inequality this yields:

    |x_n-x_m|=|x_n-x_(n+k)|=|x_n+(-x_(n+1)+x_(n+1))+(-x_(n+2)+x_(n+2))+...+(-n_(n+k-1)+x_(n+k-1)) -x_(n+k)|
    ≤|x_(n+1)-x_n|+|x_(n+2)-x_(n+1)|+...+|x_(n+k)-x_(n+k-1)|<r^n+r^(n+1)+...+r^(n+k-1)=1/r [r^(n+1)+r^(n+2)...+r^(m)]

    Using m=n+k in the last step. But the sum in the brackets is just the cauchy part of the sum of a convergent geometric series. so we can make that expression less than anything we want. In particular, we have for all ε>0, there exists and N such that for all m>n≥N:


    using this same N, we have, for all ε>0 and m>n≥N

    |x_n-x_m|≤1/r [r^(n+1)+r^(n+2)...+r^(m)]<rε/r=ε

    And now we're done=)
    Last edited: Dec 13, 2011
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