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How do you find out where a series converges

  1. Mar 6, 2005 #1
    are there certain formulae to find out where a certain infinite series converges, if it does converge

    for example
    [tex] \sum_{n=1}^\infty (\frac{3}{5})^n [/tex] certainly converges because it is between the infinite series
    [tex] \sum (1+ \frac{1}{n})^n [/tex] and the series [tex] \sum (\frac{1}{5})^n [/tex] wich both converge Since both of them converge then sum(3/5)^n must converge.

    But my question is WHERE does (3/5)^n converge??
     
    Last edited: Mar 6, 2005
  2. jcsd
  3. Mar 6, 2005 #2

    Hurkyl

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    [itex] \sum_{n=1}^\infty (3/5)^n [/itex] doesn't have any parameters, so it doesn't really make sense to ask for which values of the parameters (i.e. where) the series converges... only if.


    Now, (1 + 1/n)^n and (1/5)^n aren't series...
     
    Last edited: Mar 6, 2005
  4. Mar 6, 2005 #3
    what i meant was how do you find the explicit value of the series
    for example i know (because ive been told) that
    [tex] \sum \frac{1}{n^2} = \frac{\pi^2}{6} [/tex]
    would it be possible to that to do this series??

    note: my proof of its convergence is wrong
     
  5. Mar 6, 2005 #4

    Hurkyl

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    [itex]\sum_{n = 1}^{\infty} (1 + 1/n)^n[/itex] doesn't converge...

    Anyways, yes, your series does have a sum. It's a geometric series, so use the formula for such series.
     
  6. Mar 6, 2005 #5

    saltydog

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    Here's the thread that addressed the problem:
    https://www.physicsforums.com/showthread.php?p=470773#post470773
     
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