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Simple limit?

  1. Mar 1, 2007 #1
    simple limit?!?!?

    1. The problem statement, all variables and given/known data
    I am trying to prove if a series is convergent.
    the series is a(sub n) = (2/3)[2^n - 2^(-n)] from n=1 to infinity.

    2. Relevant equations
    Ratio test:
    the limit as n approached infinity of the absolute value of a(sub n+1) /a(sub n) equals r.
    If r is less than one the series converges.
    if r is greater than 1 the series diverges.


    3. The attempt at a solution

    lim as n approaches infinity of the absolute value of
    (2/3 * (2^(n+1) - 2 ^(-n+1)))
    (2/3 * (2^n - 2^(-n)))
    eqauls r
    the 2/3 cancel and I get
    lim as n approaches infinity of the absolute value of
    (2^(n+1) - 2 ^(-n+1))
    (2^n - 2^(-n))
    eqauls r
    this is where I get stuck. Looking at the graph of this function, I can see that the limit is 2. But I dont know how to show it. It has been a long time since I took Calc....
    PLEASE HELP
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 1, 2007 #2
    Are you talking about the series \sum a_n or a power series
    \sum a_n x^n

    or something else
    ?
     
  4. Mar 1, 2007 #3
    The series sum, sigma notation sorry I just dindt know how to put that on here
     
  5. Mar 2, 2007 #4

    HallsofIvy

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    Isn't it obvious that the sequence (2/3)(2n- 2-n) does not converge to 0?
     
  6. Mar 2, 2007 #5
    a_n = (2/3)[2^n - 2^(-n)] from n=1 to infinity.=

    [tex]a_n=\sum_{n=1}^\infty\frac{2}{3}(2^n - 2^{-n}) \lim_{1\rightarrow\infty}[/tex]

    Am I being dim here, but isn't it obvious that the total becomes arbitrarily large as n increases. Therefore it does not converge?

    2^n increases as n increases

    2^-n decreases as n increases

    thus 2/3(2^n-2^-n) becomes increasingly large, thus the sum does not converge, the 2/3 outside the brackets makes no difference to this in this case.
     
    Last edited: Mar 2, 2007
  7. Mar 2, 2007 #6
    I thought that the sequence diverged but I have to prove it not just say that it does and I dont know exactally how to show it
    Thats why I was using the ration test....but I couldnt figure out how to do the limits.
     
  8. Mar 2, 2007 #7
    right I see that but I am not sure how to "prove it" I guess I need to show that the limit is infinity but I dont know how.
     
  9. Mar 2, 2007 #8
    1. The problem statement, all variables and given/known data

    may be it will help if I included the whole problem, maybe I am not doing it correctly at all.

    Consider the sequence 1,5/2,21/4,85/8,... defined by s(sub1) =1 , a(sub2)=5/2 and a(subn)= (5/2)a(subn-1) - a(subn-2)

    A.) Show that a(subn) = (2/3)[2^n -2^(-n)] for any n greaterthan or equal to 3.

    B.) Is the series (sigma notation) from n=1 to infinity a(subn) convergent? (prove your answer)

    What I have done:
    For A.) I have proven the base case and am working on the inductive step. I know that for the inductive step, I need to show that assuming a(subn) and everything below it is true, that a(subn+1)= (5/2)a(subn) - a(subn-1) = (2/3)[2^(n+1) -2^(-n+1)] , but my algebra isnt working out. so I am still pluggin away at that.:yuck:
    For B.) I was thinking that if I show (2/3)[2^n -2^(-n) is divergent, which I can clearly see, but dont know how to prove (I was thinking of using the ratio test as quoted above)then I can conclude that Is the series (sigma notation) from n=1 to infinity a(subn) is divergent.
    so can someone help me with this equation if my reasoning is correct::uhh:

    lim as n approaches infinity of the absolute value of
    (2^(n+1) - 2 ^(-n+1))
    (2^n - 2^(-n))
    I need help showing my steps for getting the limit. :confused:
    when I plugged this equation into my calculator, I got a straight line through y=2 so I am thinking this means the limit as n approaches infinity the limit is 2, and thus by the ratio test it is divergent......
     
  10. Mar 2, 2007 #9

    cristo

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    So, you have [tex]\lim_{n\rightarrow\infty}\left|\frac{2^{n+1}-2^{1-n}}{2^n-2^{-n}}\right|[/tex], supposing you have used the ratio test correctly. Now, can you evaluate this limit? Hint: divided top and bottom by 2n
     
  11. Mar 2, 2007 #10

    HallsofIvy

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    There is a very nice theorem that says if
    [tex]\Sigma_{n=0}^\infty a_n[/tex]
    converges then {an} must converge to 0. It's contrapositive is that if {an} does not converge to 0, then [tex]\Sum_{n=0}^\infty a_n[/tex] (2/3)(2n+ 2-n) does not converge to 0. That should be easy. (In fact, it goes to infinity!)
     
  12. Mar 3, 2007 #11
    I am not sure how do do it with out getting common denomionators and such... so here is my algebra... I know that all of them have absolute value and the lim next to them but I cant figure out how to use them so I will just write the frations.

    2^(2n+2) -1
    2^(n+1)
    divided by:
    2^(2n)-1
    2^n

    which is
    (2^(2n+2)-1)(2^n)
    (2^(n+1))(2^(2n)-1)

    which is
    (2^(2n+2)-1)(2^n)
    (2^(2n+1)-2)(2^n)

    then the 2^n cancel and I am left with
    (2^(2n+2)-1)
    (2^(2n+1)-2)

    which is
    (2^(2n)*(2^2))-1
    (2^(2n)*(2))-2

    which is
    1/2??? so the limit is 1/2?
     
    Last edited: Mar 3, 2007
  13. Mar 4, 2007 #12
    as HallsofIvy says, it is quite obvious (painfully) that the series diverge.

    consider the series, 1+1+1+1+1+1+1+1+1+1+1.... obviously, it diverges, now consider the original series, every term eventually becomes bigger than 1... so...
     
    Last edited: Mar 4, 2007
  14. Mar 4, 2007 #13

    cristo

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    I'm not sure what you've done there. I said divide top and bottom by 2n to obtain [tex]\left|\frac{2^{n+1}-2^{1-n}}{2^n-2^{-n}}\right|=\left|\frac{2-2^{1-2n}}{1-2^{2n}}\right|\rightarrow ?[/tex]

    If you can see what this tends to, then you can work out your answer using the ratio test. However, Halls has given you a nice theorem which is quicker to use!
     
  15. Mar 4, 2007 #14
    Thanks all I figured it out :)
     
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