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anonymous

  • 5 years ago

please help is this absolutely or cond convergent sigma (-1)^n-1 arctan(1/n)/n^2

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  1. watchmath
    • 5 years ago
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    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.

  2. anonymous
    • 5 years ago
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    how did you come up with pi/2/x^2

  3. watchmath
    • 5 years ago
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    the range of the arctan function is between -pi/2 and pi/2

  4. anonymous
    • 5 years ago
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    and why did you put over x^2

  5. watchmath
    • 5 years ago
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    well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.

  6. anonymous
    • 5 years ago
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    so would you always compare it to pi/2 divided by denominator given in the problem

  7. anonymous
    • 5 years ago
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    for tan

  8. watchmath
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    and if there isnt anything in the denominator of the given problem you would compare it to pi/2

  10. watchmath
    • 5 years ago
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    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.

  11. anonymous
    • 5 years ago
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    okay what would you comoare arcsin (1/n) to and 1-cos1/n to

  12. watchmath
    • 5 years ago
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    the term of the series only arcsin(1/n) ?

  13. anonymous
    • 5 years ago
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    yah

  14. watchmath
    • 5 years ago
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    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...

  15. anonymous
    • 5 years ago
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    yah

  16. watchmath
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    what if it was just sin

  18. watchmath
    • 5 years ago
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    sin what?

  19. anonymous
    • 5 years ago
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    if instead of arc sin it was sin would you still make the same comparison

  20. watchmath
    • 5 years ago
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    you mean sin(1/n) ?

  21. anonymous
    • 5 years ago
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    yah

  22. watchmath
    • 5 years ago
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    well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.

  23. anonymous
    • 5 years ago
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    oh okay

  24. anonymous
    • 5 years ago
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    also if you were given sigma 1-cos(1/n) what would you compare it to

  25. watchmath
    • 5 years ago
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    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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