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Math Help - Let alpha be a real number, alpha > -1 now show that

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    Let alpha be a real number, alpha > -1 now show that

    lim n approachs infinity. the intergral from 1 to N (x^-1) dx / the intergral from 1 to N (x^alpha)dx = 0

    use the tergral from 1 to n= intergral from 1 to k + intergral form k to n
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    Re: Let alpha be a real number, alpha > -1 now show that

    Quote Originally Posted by sbacon1991 View Post
    lim n approachs infinity. the intergral from 1 to N (x^-1) dx / the intergral from 1 to N (x^alpha)dx = 0

    use the tergral from 1 to n= intergral from 1 to k + intergral form k to n

    \lim_{n\to\infty}\frac{\int_1^n \frac{1}{x}dx}{\int_1^n x^{\alpha}dx}=\lim_{n\to\infty} \frac{\ln{n}}{\frac{n^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}}=...
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    Re: Let alpha be a real number, alpha > -1 now show that

    using the de nition of the Riemann integral can you prove it for?
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    Re: Let alpha be a real number, alpha > -1 now show that

    Quote Originally Posted by sbacon1991 View Post
    using the de nition of the Riemann integral can you prove it for?
    Introducing a Riemann sum into this problem would complicate it beyond all reason. It calls for a simple application of l'Hopital's rule.
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    Re: Let alpha be a real number, alpha > -1 now show that

    is there any more info which can be given for the question?
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    Re: Let alpha be a real number, alpha > -1 now show that

    Quote Originally Posted by sbacon1991 View Post
    is there any more info which can be given for the question?
    Do you know any basic calculus?
    If not, this is a waste of time going forward.
    Last edited by Plato; November 27th 2011 at 03:27 PM.
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    Re: Let alpha be a real number, alpha > -1 now show that

    yes i do
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