Converge or diverge arctan [please move to calculus section]

[rcmango]

Homework Statement

Sorry wrong section! should be in calculus!

does this problem converge or diverge?

Homework Equations

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.

 

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

 

[rcmango]

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

 

[foxjwill]

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

 

[rcmango]

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

 

[rcmango]

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

 

[rcmango]

anyone else who can comment on what's been said here please?

 

[Feldoh]

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.