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Absolute and conditional convergence of series (1 + x/n)^n^2

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\sum[/itex] from n=1 to n=[itex]\infty[/itex] (1 + [itex]\frac{x}{n}[/itex])n2

    Determine the values of x for which the series converges absolutely, converges conditionally and diverges.


    3. The attempt at a solution

    So i tried using the root test for the absolute value of (1 + [itex]\frac{x}{n}[/itex])n2, but it was inconclusive.

    So then I tried using the ratio test, but I need to simplify this:

    |1 + [itex]\frac{x}{n+1}[/itex]|(n+1)2 / |1 + [itex]\frac{x}{n}[/itex]|n2

    and I have no idea how to simplify this.

    I don't know any other tests that we could use for this, so I'm just pretty stumped.

    Any help would be great.
     
  2. jcsd
  3. Mar 13, 2012 #2

    Dick

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    I don't think the root test is all that inconclusive. What is the nth root of your expression?
     
  4. Mar 13, 2012 #3
    So what I got was:

    nth-root(|1 + [itex]\frac{x}{n}[/itex]|n2 ) = [nth-root(|1 + [itex]\frac{x}{n}[/itex]|n )]2 = |1 + [itex]\frac{x}{n}[/itex]|2 = (1+ [itex]\frac{x}{n}[/itex])2

    So the limit of this as n goes to infinity would be 1, and that means the test is inconclusive no?

    Unless I did something wrong?
     
  5. Mar 13, 2012 #4
    Oh wait, that's not right.

    nth-root(|1 + x/n|n2) = |1 + x/n|n2/n = |1 + x/n|n

    But what does that go to when n goes to infinity?

    I don't think I know how to compute that?
     
  6. Mar 13, 2012 #5

    Dick

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    Yes, it's completely wrong. Your rules of exponents are pretty messed up. The nth root of a is a^(1/n). What's (a^(n^2))^(1/n)?
     
  7. Mar 13, 2012 #6

    Dick

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    I think you've probably seen that limit before. It's related to the exponential function. If not then you can work it by taking the log and using l'Hopital's theorem.
     
  8. Mar 13, 2012 #7
    Oh wow, I feel like an idiot now, haha.

    Yea I know that limit n[itex]\rightarrow\infty[/itex](1 + x/n)n = ex, but (1 + x/n)n doesn't necessarily equal |1 + x/n|n does it? What if x is a large negative number?

    I showed it using logarithms as you said, but I was just wondering about that?
     
  9. Mar 13, 2012 #8

    Dick

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    If x is a large negative number, then if n>|x|, 1+x/n is still a positive number. When you are doing limits you only have to worry about n very large.
     
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