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Math Help - A pair of sequence problems

  1. #1
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    A pair of sequence problems

    Does this sequence diverge or converge? If it converges, find the limit-

     (\frac{1}{n})^{(\frac{1}{n})}

    I know the harmonic sequence diverges, but I've taken natural logs of both sides and get stuck, at (1/n)(ln 1 - ln n) = ln L where L is the original limit...

    Also,

    \frac{\sqrt{n}}{2n*\csc{\sqrt{9n+4}}}

    can this be shown to converge to 0 using the squeeze theorem for sequence convergence? (note that you can move csc to the numerator as \sin{\sqrt{9n+4}})


    (both sequences are from n=1 to infinity for the natural numbers)
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  2. #2
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    Quote Originally Posted by matt.qmar View Post
    Does this sequence diverge or converge? If it converges, find the limit-

     (\frac{1}{n})^{(\frac{1}{n})}

    I know the harmonic sequence diverges, but I've taken natural logs of both sides and get stuck, at (1/n)(ln 1 - ln n) = ln L where L is the original limit...



    == I hope you already studied functions and their limits: then what you did is fine, and you need the lim (1/n) lim(1/n) when n--> oo, which can be calculated from what we about lim x ln x when x--> 0 : this limit's value is zero since x --> 0 much faster than ln x --> -oo when x --> 0.
    you can do it also using L'Hospital's rule ==. the original limit is 1.


    Also,

    \frac{\sqrt{n}}{2n*\csc{\sqrt{9n+4}}}

    can this be shown to converge to 0 using the squeeze theorem for sequence convergence? (note that you can move csc to the numerator as \sin{\sqrt{9n+4}})



    If csc is cosecant, the the function is [sin(sqrt(9n+4))]/2]*1/Sqrt(n) and
    since 1/Sqrt(n) --> 0 when n --> oo and [sin(sqrt(9n+4))]/2] is bounded the limit is zero.

    Tonio



    (both sequences are from n=1 to infinity for the natural numbers)
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