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Thread: Series help!; converge or diverge

  1. #1
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    Series help!; converge or diverge

    $\displaystyle \sum_{n=1} ^\infty (\sqrt[3]{n})^{3n} * {(\Pi-2)}^{-4n^2}
    $

    Does this converge or diverge and how?



    $\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$

    I'm leaning on converge for this one because

    the limit of $\displaystyle (1/e^x)$ = 0
    so,

    $\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$
    Converges by the Limit Comparison Test?
    Last edited by badBKO; Nov 23rd 2008 at 04:27 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badBKO View Post
    $\displaystyle \sum_{n=1} ^\infty (\sqrt[3]{n})^{3n} * {(\Pi-2)}^{-4n^2}
    $

    Does this converge or diverge and how?



    $\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$

    I'm leaning on converge for this one because

    the limit of $\displaystyle (1/e^x)$ = 0
    so,

    $\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$
    Converges by the Limit Comparison Test?
    $\displaystyle \sum_{n=1}^{\infty}\frac{n^n}{(\pi-2)^{4n^2}}$

    What about the Root test?

    $\displaystyle \begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{(\pi-2)^{4n^2}}}&=\lim_{n\to\infty}\frac{n}{(\pi-2)^{4n}}\\
    &=0<1\end{aligned}$

    The last part was gotten since the exponential function increases faster than any polynomial. So this series is convergent

    $\displaystyle \sum_{n=0}^{\infty}\sinh\left(e^{-x}\right)$

    Right idea! consider

    $\displaystyle \lim_{x\to\infty}\frac{\sinh\left(e^{-x}\right)}{e^{-x}}$

    Making the sub $\displaystyle z=e^{-x}$ gives

    $\displaystyle \begin{aligned}\lim_{z\to{0}}\frac{\sinh(z)}{z}&=\ lim_{z\to{0}}\frac{\sin(iz)}{iz}\\
    &=1\end{aligned}$

    The last part can be made with the sub $\displaystyle iz=\varphi$

    So both $\displaystyle \sum_{n=0}^{\infty}\frac{1}{e^x}$ and $\displaystyle \sum_{n=0}^{\infty}\sinh\left(\frac{1}{e^x}\right)$ share convergence.
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  3. #3
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    Thanks alot!!!!

    The first one didn't look that easy, the second one, I thought I was on the right track.
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