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anonymous
 5 years ago
please help is this absolutely or cond convergent sigma (1)^n1 arctan(1/n)/n^2
anonymous
 5 years ago
please help is this absolutely or cond convergent sigma (1)^n1 arctan(1/n)/n^2

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0The term of the absolute series is \(\arctan(1/n)/n^2<\frac{\pi/2}{n^2}\). But the series \(\sum 1/n^2\) is convergent (since it is a pseries with p=2), so the absolute series converges. Then the series is absolutely convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you come up with pi/2/x^2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0the range of the arctan function is between pi/2 and pi/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and why did you put over x^2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would you always compare it to pi/2 divided by denominator given in the problem

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and if there isnt anything in the denominator of the given problem you would compare it to pi/2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series \(\sum pi/2\) is divergent. So comparing with pi/2 doesn't give me any conclusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay what would you comoare arcsin (1/n) to and 1cos1/n to

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0the term of the series only arcsin(1/n) ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Just to make sure. So your series is \(\sum_{n=1}^\infty \arcsin(1/n)\)? I can't think how to do this right away...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0ok, arcsin is an increasing function and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1/n But the harmonic series \(\sum 1/n\) is divergent Hence \(\sum \arcsin(1/n)\) is divergent as well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what if it was just sin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if instead of arc sin it was sin would you still make the same comparison

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also if you were given sigma 1cos(1/n) what would you compare it to

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0That is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)
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