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Thread: Converges or Diverges?

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    Converges or Diverges?

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    Need to know if this converges or diverges. Could someone show me the steps ??
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    Re: Converges or Diverges?

    using the ratio test

    $L=\lim \limits_{n\to \infty}~\dfrac{\dfrac{e^{n+1}+(n+1)}{(n+1)^5}}{ \dfrac{e^n+n}{n^5}} = $

    $L=\lim \limits_{n\to \infty}~\dfrac{n^5(e^{n+1}+n+1)}{(n+1)^5(e^n+n)}$

    $L = e > 1$ and thus the series is divergent
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    Re: Converges or Diverges?

    I'm sorry I'm really bad at simplifying how did it give you e at the end ? I tried to simplify it and it never gives me e
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    Re: Converges or Diverges?

    Quote Originally Posted by steelmaste View Post
    I'm sorry I'm really bad at simplifying how did it give you e at the end ? I tried to simplify it and it never gives me e
    should be pretty clear that $\dfrac{n^5}{(n+1)^5}$ goes to 1

    should also be pretty clear that the $e^n$ term leaves the $n$ and constant terms in the dust so this becomes $\dfrac{e^{n+1}}{e^n} = e$

    you can show all that formally but that's the common sense explanation.
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    Re: Converges or Diverges?

    Quote Originally Posted by steelmaste View Post
    Click image for larger version. 

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    Need to know if this converges or diverges. Could someone show me the steps ??
    I offer another approach that uses the comparison & root test.
    $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{e^n} + n}}{{{n^5}}}} \ge \sum\limits_{n = 1}^\infty {\frac{{{e^n}}}{{{n^5}}}} $
    But
    $\displaystyle \sqrt[n]{{\frac{{{e^n}}}{{{n^5}}}}} = \frac{e}{{{{\left( {\sqrt[n]{n}} \right)}^5}}} \to e > 1$
    This shows divergence.
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    Re: Converges or Diverges?

    Thank you ! this approach seems a lot more simpler, however I fail to understand how to make the bottom part (nsquarerootn)^5 go to 1 for it to be e/1. any tricks to simplify I could use? I pretty much tried to make it (n^(1/n))^5 but still dont see how it goes to 1, since my teacher said we cant just assume that infinity^(0) gives 1.
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    Re: Converges or Diverges?

    Quote Originally Posted by steelmaste View Post
    Thank you ! this approach seems a lot more simpler, however I fail to understand how to make the bottom part (nsquarerootn)^5 go to 1 for it to be e/1. any tricks to simplify I could use? I pretty much tried to make it (n^(1/n))^5 but still dont see how it goes to 1, since my teacher said we cant just assume that infinity^(0) gives 1.
    If you do not know that $\displaystyle \sqrt[n]{n} \to 1$ then you are not able to use the root test.
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    Re: Converges or Diverges?

    Quote Originally Posted by steelmaste View Post
    Thank you ! this approach seems a lot more simpler, however I fail to understand how to make the bottom part (nsquarerootn)^5 go to 1 for it to be e/1. any tricks to simplify I could use? I pretty much tried to make it (n^(1/n))^5 but still dont see how it goes to 1, since my teacher said we cant just assume that infinity^(0) gives 1.
    Let $y = n^{\frac 1 n}$. Then $\ln y =\frac{\ln n}{n} \to 0$. So $y\to 1$.
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