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Lim convergent sequence

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Proove rigorously that if (a[tex]_{n}[/tex] is a real convergent sequence with lim[tex]_{n\rightarrow \infty}[/tex] a[tex]_{n}[/tex] = a and for each n=[tex]\in[/tex] N, a[tex]_{n}[/tex] < 6, then a [tex]\leq[/tex] 6
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    Let [tex]\epsilon[/tex] > 0 we need to find n[tex]_{0}[/tex] [tex]\in[/tex] N such that

    [tex]\left\|[/tex] a[tex]_{n}[/tex] - a[tex]\left\|[/tex] < [tex]\epsilon[/tex] [tex]\forall[/tex] n [tex]\geq[/tex] n [tex]_{0}[/tex], n[tex]_{0}[/tex] [tex]\in[/tex] N

    but a[tex]_{n}[/tex] < 6
    so

    [tex]\left\|[/tex] 6 - a[tex]\left\|[/tex] < [tex]\epsilon[/tex]

    then

    a < 6 - [tex]\epsilon[/tex] and [tex]\epsilon[/tex] > 0

    so a [tex]\leq[/tex] 6

    i think i've done this right, just by using the definition of a limit. Could anyone tell me if this is looking ok?

    (Also i cant seem to get the sub script working, it always makes them go up instead of down, any ideas anyone?)

    Thanks a million
     
    Last edited: Dec 7, 2009
  2. jcsd
  3. Dec 7, 2009 #2

    statdad

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    Homework Helper

    No, you're off track. You know that the sequence converges, so no matter what [tex] \epsilon [/tex] is given you can make

    [tex]
    |a_n - a| < \epsilon
    [/tex]

    if n is large enough.
    But this:


    doesn't follow.

    Try assuming [tex] a > 6 [/tex] and see if you can reach some contradiction. (Hint: if [tex] a > 6 [/tex] then [tex] a - 6 > 0 [/tex].)
     
  4. Dec 8, 2009 #3
    thanks for getting back to me.

    ok let me see if i have this right now.

    first since the sequence is convergent [tex]\epsilon[/tex] > 0. then set a > 6 so

    Since the sequence is convergent a[tex]_{}n[/tex] - a < [tex]\epsilon[/tex] or a < a[tex]_{}n[/tex] - [tex]\epsilon[/tex] but if a > 6 then 6 - [tex]\epsilon[/tex] > a > 6 which cant be true so a muxt be < or = 6
     
  5. Dec 8, 2009 #4

    statdad

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    Homework Helper

    Not quite . My point was that since [tex] a - 6 >0 [/tex] it would be a possibility for choice as [tex] \epsilon [/tex]
     
  6. Dec 8, 2009 #5
    thanks for all the help,

    do you mean a - 6 > 0 and an -a < [tex]\epsilon[/tex] then an - a < a - 6 ?
     
  7. Dec 8, 2009 #6
    I must be still lookin at this wrong, cant seem to figure it out
     
  8. Dec 8, 2009 #7
    Think of it intuitively first.

    If [tex] a_n < 6 [/tex] for all n, and you want to prove that [tex] a \le 6 [/tex], then assume the opposite. Assume that [tex] a > 6 [/tex]. But since [tex] a_n \rightarrow a [/tex] then that means we can make [tex] a_n [/tex] as close to a as we want, provided that n is taken large enough, right?

    Do you see how if a > 6 then there is no way we can possibly make [tex] a_n [/tex] as close to a as we want, no matter how large n is? Because for all n, [tex] a_n < 6 [/tex] ?

    Make this into a rigorous argument now.
     
  9. Dec 8, 2009 #8
    thanks a mill for replying

    I can see that if an < 6 and a > 6 but an [tex]\rightarrow[/tex] a so they cant converge as they are both on opposite sides of 6 so to speak, but im just not sure how to go about prooving this rigourously
     
  10. Dec 8, 2009 #9
    Assume a > 6. Since [tex] a_n < 6 < a [/tex] for all n, we have [tex] a - a_n > 0 [/tex]. So [tex] |a_n - a| = a - a_n [/tex].

    Now, we know that for every positive [tex] \epsilon [/tex] , [tex] a - a_n < \epsilon [/tex] provided that n is large enough. If a > 6 then a - 6 > 0, and so we can take [tex] \epsilon = a - 6 [/tex]. Try that out and see what happens.
     
  11. Dec 8, 2009 #10
    the first part of this is grand. the second part, im not following are you getting it from the def of a limit [tex]\left|[/tex]an-a [tex]\left|[/tex] < [tex]\epsilon[/tex]
     
  12. Dec 8, 2009 #11
    In the part you quoted, I'm just making it clear that |a_n - a| = a - a_n.

    I'd rather get rid of the absolute value bars.

    The rest of my post uses the definition of a limit.
     
  13. Dec 8, 2009 #12
    sorryi'm not getting this, is this just because a>6>a_n
     
  14. Dec 8, 2009 #13
    Yes. You know that |a_n - a| = |a - a_n|, right? but a - a_n > 0 anyway since a > a_n, so we can just drop the absolute value bars and write a - a_n
     
  15. Dec 8, 2009 #14
    Damn this just isnt working for me- [tex]\epsilon = a- 6[/tex] but [tex]\epsilon>0[/tex] so a-6> 0 or a>6 which is what i'm trying to disprove!
     
  16. Dec 8, 2009 #15
    a-an < [tex]\epsilon[/tex] and a - 6 = [tex]\epsilon[/tex]

    then a-an < a - 6

    which works out to an < 6 which is true???
     
  17. Dec 8, 2009 #16

    statdad

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    Homework Helper

    [tex]
    a - a_n < a - 6
    [/tex]

    leads to

    [tex]
    a_n < 6
    [/tex]?

    Check your signs again.
     
  18. Dec 8, 2009 #17
    damn it, sorry bout that. thanks a million for the help
     
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