# Converge or diverge arctan [please move to calculus section]

by rcmango
Tags: arctan, calculus, converge, diverge
 P: 234 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.
 P: 355 Is you're question whether $$\lim_{x\rightarrow \infty}\arctan{x}$$ exists? My first instinct is to try for a series representation, but I'm not sure. Edit: Come to think of it, by definition, the range of $$\arctan{x}$$ is $$(-\frac{\pi}{2},\frac{\pi}{2})$$ while its domain is $$(-\infty, \infty)$$, so, not only does $$\lim_{x\rightarrow \infty}\arctan{x}$$ exist, it equals $$\frac{\pi}{2}$$.
 P: 234 Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?
 P: 355 Converge or diverge arctan [please move to calculus section] I don't know (>_<), but it's kind of not something that needs to be proven since it's part of the definition.
 P: 234 So if i show the definition as work, would you believe that to be enough? thanks for all the help.
 P: 234 also, could i show this using the squeeze theorem with bounds -pi/2 and upper bound pi/2
 P: 234 anyone else who can comment on whats been said here please?
P: 1,345
 Quote by rcmango Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?
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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