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Thread: Series convergence test... double square root?

  1. #1
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    Series convergence test... double square root?

    Hi guys, new to the forum. I had this question on a test in Calculus, and it really stumped me.

    [math]\sum_{n=2}^{\infty}\frac{1}{\sqrt{n*\sqrt{n-1}}}[\math]

    I don't know if i wrote that correctly, but it's essentially a series with a denominator consisting of sqrt(n(sqrt(n-1))), starting at 2 and going to infinity. The instructions for the question is to determine whether the series converges or diverges, and to state the method used. I tried splitting the denominator into n's to the powers of -1/2 and -1/4, then using the partial fractions. But my attempt was incorrect.

    Does anyone have any idea how to approach this problem? I really thought I was onto something with the fractional exponents.

    (Another image, just in case.)


    (EDIT: yup, totally messed up the latex code. Sorry about that.)
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  2. #2
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    Re: Series convergence test... double square root?

    Just compare this series with the harmonic series.

    $n>1 \Rightarrow \dfrac{1}{\sqrt{n\sqrt{n-1}}} > \dfrac 1 n$

    $\sum \dfrac 1 n$ is known to diverge so this series must as well.
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  3. #3
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    Re: Series convergence test... double square root?

    Thank you for the explanation! Turns out I was thinking too much about it after all.
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