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Trouble with a converging sequence.

  1. Apr 2, 2004 #1
    Trouble with a converging sequence.......

    Having a numpty moment here with a proof, so any help would be great! Thanks!

    If (xn) is a convergent sequence with a non-zero limit, prove that there exists a positive integer N such that mod(xn)>amod(limxn) where n tends to infinity for the lim part there and 0<a<1

    So I let the limxn bit just be l to make life a bit easrer (hope it's okay to do that) and I statred trying to prove things using the triangle inequlity etc, but it just won't seem to work. For example-
    mod(xn)=mod(xn - l + l)<mod(xn-l) + mod(l). but I couldn't get anything from this.

    or mod(xn)=mod(xn -l +l)> mod(mod(xn+l) - mod(l))=mod(mod(l)- mod(xn+l))> mod(l)- mod(xn+l) > amod(l)-mod(xn+l) However, I can't get anything from this either. Slightly bamboozled here which is a bit worrying cosnidering it's not exactly rocket scienece being required to get it. :frown:
  2. jcsd
  3. Apr 2, 2004 #2


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    What do you mean by "mod"?
  4. Apr 2, 2004 #3

    matt grime

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    Ok, I'm in a foul mood (and you gender is to blame! buit hopefully I can work through that), and i think that you want to show that for all a with 0<a<1 that if x(n) converges to x that there is an N such that for all n>N (for some choice of N dependeing on a)

    |x(n)| > a|x(n)|

    correct? as long as x is not zero.

    Well, we may suppose that x(n) is a sequrnce of positive terms, just so we can drop the mod signs.

    can you relate the ax(n) < x(n) thing to an epsilon situation which is all that is going on? (you're replacing absolute errors, with relative errors, that's all).
  5. Apr 2, 2004 #4


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    There's a real notation problem here:

    Let's say we have [tex]x:\mathbb{N} \rightarrow \mathbb{R}[/tex] a convergent sequence. Then we say that [tex]x(n)[/tex] is the [tex]n^{th}[/tex] element of [tex]x[/tex].

    Now, for [tex]1>a>0[/tex]
    [tex]|x(n)| > a |x(n)| [/tex] is trivially true for all n.

    If you mean that given [tex]1 > \epsilon > 0 [/tex] there exists [tex]N[/tex] so that
    [tex]N<n \rightarrow |x(N)| > |x(n)|[/tex], that is also not generally true.
    (For an example sequence, consider [tex]x(n) = 1-2^{-n}[/tex] which clearly converges to [tex]1[/tex] and is clearly monotone positive.)

    If this discussion is about a convergent series [tex]\sum x_n[/tex] then you know that [tex]\lim_{x\rightarrow\infty} x_n = 0[/tex] and you can prove that there are infinitely many [tex]N[/tex] so that [tex]n>N \rightarrow |x_N|>|x_n| or |x_n|=0[/tex], but the limit that is suggested above is quite trivial.

    (Of course, Claire might be trying to prove that [tex]\lim_{x\rightarrow\infty} x_n = 0[/tex] )
  6. Apr 3, 2004 #5

    matt grime

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    correction, i mean x(n)>ax not ax(n). the question isn't about series, it can't be as the limit of x(n) is not zero.
  7. Apr 3, 2004 #6
    Sorry to hear about woman-trouble Matt, but we aren't all evil. :rolleyes: Plus there are many members of your species who are't so delightful either (not you of course, I hope, but I've had my fair share of fun at uni this year which has been a bit on an eye-opener let's just say!). Oh well, at least count yourself lucky that you aren't currently being accused of having an affair with a married man (unless you are and there's something you're not telling us! :wink: ).

    Anyoo, I tried to do stuff with epsilon here but it just won't work. Feeling very blonde today (despite being dark-haired!). Like if we have mod(xn - x)<e then we can't like shove mod(xn-ax) less than that. And I can see where you're coming from about dropping the mod signs as you did above but I'm not really sure ho to apply it to this situation, apart from if you put mod(xn)-mod(x)< mod(xn - x) but even then I'm not sure where you'd take it from here since you're going to be taking away a smaller number (an as opposed to x).
  8. Apr 3, 2004 #7

    matt grime

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    Let's just do part of manipulation hope you'll et me shove in a new letter:

    y-ax = y-x+(1-a)x

    does that help you? Pick N such that mod(x(n)-x) < (1-a)x then remember that |z|<w is the same as saying -w<z<w
  9. Apr 3, 2004 #8
    Okay so if I choose mod(xn-x)<(1-a)x, I presume this is kinda like having e since it's a real number greater than zero (since we're presuming that the limit is positive here, right?).

    So this means we have -(1-a)x<xn-x<(1-a)x

    Is this where you bring in the manipulation that you mentioned with the y's in it (which I presume you use xn instead of here, right....).

    From -(1-a)x<xn-x you can get -x + ax<xn-x so the x's would cancel leaving you with ax<xn but I don't think you can od that somehow... :confused:
  10. Apr 3, 2004 #9

    matt grime

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    why can't you do it? That's how it works. Remember we've replaced the original x(n) by |x(n)| at all points and this converges to |x| a nonzero number.
  11. Apr 4, 2004 #10
    Oh I kinda see now. So that when we write mod(xn-x) normally we actually have mod(mod(xn)-(mod(x)) since we're loking at the modulus of the sequence as opposed to just the sequence (xn). So then do you put this <(1-a)mod(x) so that you then get


    so you then get-


    which then gives you-


    is this right?
    Last edited: Apr 4, 2004
  12. Apr 4, 2004 #11
    Oh I've got one more thing to pester you brain with (I seem to be very good at that, one of the few things I am actually good at! :tongue: ).

    If we have (xn) which is a convergent sequence with limit x, and suppose than xn>=0 for each natural n, prove that sqrt(xn) tends to sqrt(l) as n tends to infinity.

    I started it so we had mod(xn-x)<e where e is any real number>0

    Then I got that mod((sqrt(xn)-sqrt(x))(sqrt(xn)+sqrt(x))=mod(xn-x)

    so then we'd have mod((sqrt(xn)-sqrt(x))(sqrt(xn)+sqrt(x))<e

    I think we're wanting to show that mod(sqrt(xn)-sqrt(x))<e but I'm not sure how to get that from what I've got aove.

    I thought I could have done something along the lines of

    mod(sqrt(xn)-sqrt(x))<mod((sqrt(xn)-sqrt(x))(sqrt(xn)+sqrt(x)) but the sqrt(xn)+sqrt(x) could be less than 1 so I don't think you can do it that way. Am I going about this in the correct way or not? Thanks. :smile:
  13. Apr 4, 2004 #12

    matt grime

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    Firstly we know that x(n) is bounded some number M, and we know x is fixed. so (sqrt(x_n)+sqrtx) < (sqrtM+sqrtx) call this constant k, ok?
    let me set y_n = sqrt(x_n) and y=sqrtx

    so we must show that, given e>0 we can find m such that |y_n-y| <e

    Now youve got the trick right:

    |x(n)-x| = |y_n-y||y_n+y|,

    so |y_n-y| < |x_n-x|/k

  14. Apr 4, 2004 #13
    Hmm I don't know how you get the second last line there because I thought that the sign would be the other way round because you're dividing by a bigger number than mod(yn + y). I've probably just done something silly!
  15. Apr 4, 2004 #14

    matt grime

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    Ok, perhaps I've made a mistake, let me think....

    Can I add that you've picked the epsilon at the wrong time. Remember e is just a label and has no intrinsic meaning other than 'some positive real number'
    Last edited: Apr 4, 2004
  16. Apr 4, 2004 #15
    Is your k here just sqrtM + sqrtx?

    And please excuse my stupidity here. Let's just hope my brainwakes up before the exams (it did at Crimbo which was quite miraculous to be honest)).

    Btw, I take it you're not coming here for the Maths thing then?
    Last edited: Apr 4, 2004
  17. Apr 4, 2004 #16
    Hm whereabouts should it be then? I just chose it where it was because of what they'd given it in the question.
  18. Apr 4, 2004 #17

    matt grime

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    Right, here's a particular example when lim(x_n)=0 to then show you how to pick e's and so on(letting y_n be the sqrt of x_n): we show that y_n is a null sequence too, firstly given f>0 then for all n sufficiently large 0<= x_n<= f, agreed? So 0<y_n < sqrt(f). So given e>0 let f=e^2.

    for the general case you'll need to think a little harder, but if you correct what I wrote above (bounding x_n+x below, by x, say) should do it.

    More generally, if we know x_n is convergent and y_n is related to x_n somehow then if we want to show |y_n-y|<e for some e we may have to (will have to) find a way to make |x_n-x| < f(e) some other positive real number depending on e somehow for all n sufficiently large.
    Last edited: Apr 4, 2004
  19. Apr 4, 2004 #18
    Rightio, this is all kinda new to me cos we've never had to think like this (anything we've done has really just been with shoving e in at the start and working from there and generally they only take a line or something, we've never gad to choose an f that depends on e). I mean I can see why you're doing it but I wish we'd been told about it. Or maybe it's just something we should have known.

    Just to make sure I'm going about this okay before I start faffing around with e's and f(e)'s, just want to see if this is okay-

    we have (xn) bounded below by M. Therefore sqrt(xn) + sqrt(x)>sqrt(x)+sqrt(M) sine M<xn or would this need to be greater than or equal to?

    Then you'd have mod(yn -y)<mod(xn - x)/ k where k is sqrt(x)+sqrt(M)
  20. Apr 4, 2004 #19

    matt grime

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    I think you were right and I'd made an error in my < or > signs.

    |x_n-x| = |y_n-y||y_n+y| > |y_n-y| |y| if y is not zero, remember y=sqrt(x)

    so, given e>0 for n sufficiently large |x_n-x| < e|y|

    ie for n sufficiently large

    e|y| > |yn-y||y|

    cancel the y's

    It is all pointless book keeping but in the simple cases it's good to see what's going on. What we are aiming to do is make something arbitrarily small, and in that case being less than e is just as good as being less than 3e or e^2.

    Here's an example:

    suppose x_n tends to x, show 2x_n tends to 2x

    well, case 1
    given e>0 we know that |x_n-x| < e/2 for n sufficently large, and thus |2x_n-2x| < 2e/2 = e

    given e>0 we know that |x_n-x| < e for n sufficiently large, and thus |2x_n-2x| <2e

    both of which are perfectly good ways of proving the same fact. It's just that to begin with, until you get more familiar, it is often better, though time consuming to show it is exactly less than e on the nose. It is also completely pointless, in my opinion, as long as you end up with something less than some object, dpeending on e say, that can be made arbitrarily small itself (2e can be made samller than any given f by picking e< f/2) then you are done. I've often seen fantastic proofs of fantastic statements get bogged down as the leacturer picks some thing less than (r^4)/20 so that in the end he can add together r/3 three times and get an r out exactly just so he can then let r tend to zero!
  21. Apr 4, 2004 #20
    Ah I see what you mean now! I guess I just got a little too used to having a easy time with the e's at the start of the question! So pretty much as long as we end up with the e at the end we can have the initial expression less than any real number that's greater than zero, in this instance emod(y). I hope I'm making sense there (I usually don't!).

    Thanks for explaing all that to me, it's easier to understand some of the concepts when you can actually discuss hem with someone as opposed to just reading off lecture notes.
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