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Converge or diverge arctan [please move to calculus section] 
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#1
Jan1608, 08:37 PM

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. 


#2
Jan1608, 08:51 PM

P: 355

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.
Edit: 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]. 


#3
Jan1608, 10:39 PM

P: 234

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



#4
Jan1608, 10:57 PM

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.



#5
Jan1608, 11:32 PM

P: 234

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



#6
Jan1608, 11:37 PM

P: 234

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



#7
Jan2008, 07:23 PM

P: 234

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



#8
Jan2008, 07:32 PM

P: 1,345

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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