Converge or diverge arctan [please move to calculus section]

1. Jan 16, 2008

rcmango

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. Jan 16, 2008

foxjwill

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 $$\frac{\pi}{2}$$.

Last edited: Jan 16, 2008
3. Jan 16, 2008

rcmango

Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?

4. Jan 16, 2008

foxjwill

I don't know (>_<), but it's kind of not something that needs to be proven since it's part of the definition.

5. Jan 16, 2008

rcmango

So if i show the definition as work, would you believe that to be enough? thanks for all the help.

6. Jan 16, 2008

rcmango

also, could i show this using the squeeze theorem with bounds -pi/2 and upper bound pi/2

7. Jan 20, 2008

rcmango

anyone else who can comment on whats been said here please?

8. Jan 20, 2008

Feldoh

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.