anonymous
  • anonymous
please help is this absolutely or cond convergent sigma (-1)^n-1 arctan(1/n)/n^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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watchmath
  • watchmath
The 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 p-series with p=2), so the absolute series converges. Then the series is absolutely convergent.
anonymous
  • anonymous
how did you come up with pi/2/x^2
watchmath
  • watchmath
the range of the arctan function is between -pi/2 and pi/2

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anonymous
  • anonymous
and why did you put over x^2
watchmath
  • watchmath
well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.
anonymous
  • anonymous
so would you always compare it to pi/2 divided by denominator given in the problem
anonymous
  • anonymous
for tan
watchmath
  • watchmath
I 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
  • anonymous
and if there isnt anything in the denominator of the given problem you would compare it to pi/2
watchmath
  • watchmath
No, 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
  • anonymous
okay what would you comoare arcsin (1/n) to and 1-cos1/n to
watchmath
  • watchmath
the term of the series only arcsin(1/n) ?
anonymous
  • anonymous
yah
watchmath
  • watchmath
Just to make sure. So your series is \(\sum_{n=1}^\infty \arcsin(1/n)\)? I can't think how to do this right away...
anonymous
  • anonymous
yah
watchmath
  • watchmath
ok, 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
  • anonymous
what if it was just sin
watchmath
  • watchmath
sin what?
anonymous
  • anonymous
if instead of arc sin it was sin would you still make the same comparison
watchmath
  • watchmath
you mean sin(1/n) ?
anonymous
  • anonymous
yah
watchmath
  • watchmath
well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
also if you were given sigma 1-cos(1/n) what would you compare it to
watchmath
  • watchmath
That 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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