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Question about a limit

  1. Jul 15, 2012 #1
    This problem is taken from the visual calculus archives:
    http://archives.math.utk.edu/visual.calculus/6/series.15/index.html

    Determine whether the series is convergent:

    infinity
    Ʃ (1-2^n)/(1+2^n)
    n=0

    Checking this on WolframAlpha I see that the limit of the series is -1, so the series is divergent by the Nth term test. However, I don't understand why the limit is -1.

    The explanation given by WolframAlpha seems to rely on some algebraic manipulation which does not immediately jump out at me. Is the process by which WolframAlpha arrived at the answer the most efficient? Should I have seen that? Or is there a better way?

    http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+(1-2^n)/(1+2^n)
     
  2. jcsd
  3. Jul 15, 2012 #2

    sharks

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    [tex]\lim_{n\to{\infty}} \frac{1-2^n}{1+2^n}=\lim_{n\to{\infty}} \frac{2^n(1/2^n-1)}{2^n(1/2^n+1)}[/tex] Can you continue from here?
     
  4. Jul 15, 2012 #3
    Sure. Using rules for limits...

    [tex]\lim_{n\to{\infty}} \frac{2^n(1/2^n-1)}{2^n(1/2^n+1)} = \lim_{n\to{\infty}} \frac{2^n}{2^n}\times\lim_{n\to{\infty}}\frac{(1/2)^n-1}{(1/2)^n+1} \\=\lim_{n\to{\infty}} \frac{2^n}{2^n}\times\left [\lim_{n\to{\infty}}\frac{(1/2)^n}{(1/2)^n} + \lim_{n\to{\infty}}\frac{-1}{+1}\right ] [/tex]

    The first term goes to 0, (inf/inf), which should make the other two terms go to zero regardless of what they come out to be. The third term must be where the final answer is coming from, but why?
     
  5. Jul 15, 2012 #4

    HallsofIvy

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    "The first term goes to 0"? I'm not sure which you are referring to as the "first term" but both [itex]2^n/2^n[/itex] and [itex](1/2)^n/(1/2)^n[/itex] are equal to 1 for all x and so go to 1. Of course, -1/1= -1 for all n so its limit is 1.
     
  6. Jul 15, 2012 #5
    You're right. I was looking at [tex]2^n/2^n[/tex] and thinking that the result would be [tex]\infty/\infty[/tex], or undefined or zero. Perhaps it is inappropriate to think of an undefined term as 0?

    Either way I was wrong, because it isn't [tex]\infty/\infty[/tex] but rather [tex]2^{\infty}/2^{\infty}[/tex] which is certainly one.
     
  7. Jul 15, 2012 #6

    HallsofIvy

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    Unfortunately, you are still wrong and seem to have a completely wrong idea about limits.
    The limit is NOT [itex]\infty/\infty[/itex] nor is it [itex]2^\infty/2^\infty[/itex] since neither "[itex]\infty[/itex]" nor "[itex]2^\infty[/itex]" since neither of those is a number that can be a limit.
     
  8. Jul 15, 2012 #7
    Why is the limit 1?
     
  9. Jul 15, 2012 #8
    As n approaches infinity 2^n/2^n becomes larger and larger, but [tex]lim_{n\to\infty}[/tex]2^n/2^n will always be 1 because they grow larger and larger together.

    1/1 = 1
    32/32=1
    4096/4096=1
    2^2,000,000/2^2,000,000=1

    If you think of it as f(x)=2^x/2^x, no matter what we put in as x (assuming x>0) we will never get anything >1

    EDIT: clarify that I am talking about the limit
     
  10. Jul 15, 2012 #9
    I think he meant -1
     
  11. Jul 15, 2012 #10
    I'm new to this so I'm probably wrong, but here are my two cents:

    I'm not sure splitting the limit like that works.

    limit (1/2^n-1)/(1/2^n+1) doesn't equal limit (1/2^n)/(1/2^n) + limit (-1/1).

    (x+a)/(x+b) doesn't equal x/x + a/b, it equals x/(x+b) + a/(x+b).

    (also if your way is correct the limits end up being 1 X [1-1] which doesn't equal -1 which you said you know is the correct answer)

    If you factor out the 2^n you're left with

    (1/2^n-1)/(1/2^n+1) which if you evaluate as a whole piece and don't split it, gives you -1.

    Like I said, probably wrong.
     
  12. Jul 15, 2012 #11
    dividing both nominator and denominator by 2n yields [itex]\lim _{n\to\infty} \frac{\frac{1}{2^n}-1}{\frac{1}{2^n}+1}[/itex], which is trivial to evaluate, isn't it?

     
  13. Jul 15, 2012 #12

    HallsofIvy

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    That was a typo. I mean -1, of course.
     
  14. Jul 15, 2012 #13
    You're right, I can't decompose that fraction that way. Take a look at this:

    [tex]\lim_{n \to \infty} \frac{2^n}{1+2^n}=\lim_{n \to \infty} \frac{1}{1+2^n}-\lim_{n \to \infty} \frac{2^n}{1+2^n}[/tex]

    [tex]\lim_{n \to \infty} \frac{1}{1+2^n}=0[/tex]

    Leaving us with

    [tex]-\lim_{n \to \infty} \frac{2^n}{1+2^n} = \frac{\lim_{n\to\infty}2^n}{\lim_{n\to\infty}1+2^n} = -1[/tex]

    I may have made a mistake, but I feel good about this answer, but could someone tell me if there is an intermediate step in writing the last two steps? Because:

    [tex]\lim_{n\to\infty}2^n = \infty\text{ and}\lim_{n\to\infty}1+2^n=\infty[/tex]
    [tex]\text{but} \frac{\infty}{\infty}\neq 1[/tex]

    See what I mean?
     
  15. Jul 15, 2012 #14
    Yes. That is very easy and very intuitive, at least in my mind.

    Thanks, by the way!
     
    Last edited: Jul 15, 2012
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