Converge or diverge arctan [please move to calculus section]

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Sorry wrong section!! should be in calculus!!

    does this problem converge or diverge?

    2. Relevant equations

    3. The attempt at a solution

    Not sure where to start, i've never seen a series arctan problem before, is there a way to switch it around to look like i want it too, maybe with sin or cos. please help.
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2
    Is you're question whether [tex]\lim_{x\rightarrow \infty}\arctan{x}[/tex] exists? My first instinct is to try for a series representation, but I'm not sure.

    Come to think of it, by definition, the range of [tex]\arctan{x}[/tex] is [tex](-\frac{\pi}{2},\frac{\pi}{2})[/tex] while its domain is [tex](-\infty, \infty)[/tex], so, not only does [tex]\lim_{x\rightarrow \infty}\arctan{x}[/tex] exist, it equals [tex]\frac{\pi}{2}[/tex].
    Last edited: Jan 16, 2008
  4. Jan 16, 2008 #3
    Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?
  5. Jan 16, 2008 #4
    I don't know (>_<), but it's kind of not something that needs to be proven since it's part of the definition.
  6. Jan 16, 2008 #5
    So if i show the definition as work, would you believe that to be enough? thanks for all the help.
  7. Jan 16, 2008 #6
    also, could i show this using the squeeze theorem with bounds -pi/2 and upper bound pi/2
  8. Jan 20, 2008 #7
    anyone else who can comment on whats been said here please?
  9. Jan 20, 2008 #8
    You could.

    arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - ... (-1 < x < 1) is the maclaurin series for it.
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