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Math Help - Series - determine if this series is convergent/divergent

  1. #1
    Newbie Marbel's Avatar
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    Exclamation Series - determine if this series is convergent/divergent

    Hi, I need help with this series:

    from n=2 to infinity, n/[ln(n)]^n

    inf
    Σ ____n_____
    n=2 [ln(n)]^n

    I tried the ratio test and when I went to take L'Hopital's I got into a big mess, can someone show me the solution?

    Thank-you!
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  2. #2
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    Quote Originally Posted by Marbel View Post
    Hi, I need help with this series:

    from n=2 to infinity, n/[ln(n)]^n

    inf
    Σ ____n_____
    n=2 [ln(n)]^n

    I tried the ratio test and when I went to take L'Hopital's I got into a big mess, can someone show me the solution?

    Thank-you!

    The n-th roots test gives you the answer at once...or you can also try the condensation test.

    Tonio
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  3. #3
    Newbie Marbel's Avatar
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    Okay thanks!
    I went to a TA a last week and he confused me completely (had no idea what he was doing.. I ended up explaining the material to him). I can see now how it works, I forgot about my limit laws and was wondering what the heck to do with an indeterminate type of ∞^0/∞. This also was what got me into a huge mess with the ratio test...
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  4. #4
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    You can use the ratio test
    \lim_{n\rightarrow \infty}\left | \frac{n+1}{\left (\ln (n+1)  \right )^{n+1}}\cdot\frac{(\ln n)^{n}}{n}   \right |

    Expand the n+1 exponent
    \lim_{n\rightarrow \infty} \frac{n+1}{\left (\ln (n+1)  \right )^{n}\cdot \ln (n+1)}\cdot\frac{(\ln n)^{n}}{n}

    Divide stuff so that using limit laws so it is easier to take the limit
    \lim_{n\rightarrow \infty }\frac{n+1}{n}\cdot \lim_{n\rightarrow \infty }\frac{\left (\ln n  \right )^n}{\left ( \ln (n+1) \right )^n}\cdot \lim_{n\rightarrow \infty }\frac{1}{\ln (n+1)}

    The first limit goes to 1 and the third limit goes to zero, but the second needs more work so raise the whole thing to the shared exponent
    \lim_{n\rightarrow \infty }\left (\frac{\ln n}{\ln (n+1)}  \right )^{n}\cdot 0

    you can move the limit inside because of the power limit law
    \left (\lim_{n\rightarrow \infty }\frac{\ln n}{\ln (n+1)}  \right )^{n}\cdot 0

    now you can use l'hospital's rule on the inside
    \left (\lim_{n\rightarrow \infty }\frac{\frac{1}{n}}{\frac{1}{n+1}}  \right )^{n}\cdot 0

    with inverting and flipping you get this...
    \left (\lim_{n\rightarrow \infty }\frac{n+1}{n}  \right )^{n}\cdot 0<br />

    the limit goes to 1 and multiplied by zero gives zero which is less than 1 so by the ratio test...
    \sum_{n=2}^{\infty }\frac{n}{\left (\ln n  \right )^{n}} converges
    whoo, that was a mouthful
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    Quote Originally Posted by Slyprince View Post
    You can use the ratio test
    \lim_{n\rightarrow \infty}\left | \frac{n+1}{\left (\ln (n+1) \right )^{n+1}}\cdot\frac{(\ln n)^{n}}{n} \right |

    Expand the n+1 exponent
    \lim_{n\rightarrow \infty} \frac{n+1}{\left (\ln (n+1) \right )^{n}\cdot \ln (n+1)}\cdot\frac{(\ln n)^{n}}{n}

    Divide stuff so that using limit laws so it is easier to take the limit
    \lim_{n\rightarrow \infty }\frac{n+1}{n}\cdot \lim_{n\rightarrow \infty }\frac{\left (\ln n \right )^n}{\left ( \ln (n+1) \right )^n}\cdot \lim_{n\rightarrow \infty }\frac{1}{\ln (n+1)}

    The first limit goes to 1 and the third limit goes to zero, but the second needs more work so raise the whole thing to the shared exponent
    \lim_{n\rightarrow \infty }\left (\frac{\ln n}{\ln (n+1)} \right )^{n}\cdot 0

    you can move the limit inside because of the power limit law
    \left (\lim_{n\rightarrow \infty }\frac{\ln n}{\ln (n+1)} \right )^{n}\cdot 0


    Uh? What is that rule? And anyway no, you cannot move the limit "inside" when ALSO the power depends on the variable that is in the limit! If you

    could then you'd get e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\l  eft(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\ri  ght)^n=(1)^n=1 , which of course is wrong.

    In this case the limit certainly is 1, but the way to get there is wrong.


    Tonio



    now you can use l'hospital's rule on the inside
    \left (\lim_{n\rightarrow \infty }\frac{\frac{1}{n}}{\frac{1}{n+1}} \right )^{n}\cdot 0

    with inverting and flipping you get this...
    \left (\lim_{n\rightarrow \infty }\frac{n+1}{n} \right )^{n}\cdot 0
    the limit goes to 1 and multiplied by zero gives zero which is less than 1 so by the ratio test...

    math]\sum_{n=2}^{\infty }\frac{n}{\left (\ln n \right )^{n}}[/tex] converges
    whoo, that was a mouthful
    .
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  6. #6
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    Quote Originally Posted by Marbel View Post
    Hi, I need help with this series:

    from n=2 to infinity, n/[ln(n)]^n

    inf
    Σ ____n_____
    n=2 [ln(n)]^n

    I tried the ratio test and when I went to take L'Hopital's I got into a big mess, can someone show me the solution?

    Thank-you!
    If, rather than applying a "black box" convergence test, you would like to "feel" why this series converges, you can note that (\ln n)^n\geq 4^n as soon as n>e^4, hence \frac{n}{(\ln n)^n}\leq \frac{n}{4^n}. So you see that the general term converges to 0 very quickly. Since n<2^n for large n (in fact, for all n), we have 0\leq\frac{n}{(\ln n)^n}\leq \frac{1}{2^n} for large n. Since the geometric series \sum_n\frac{1}{2^n} is well known to converge, your series converges.
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  7. #7
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    Thanks for pointing out my mistake, Tonio. I'll be sure not to make it again. For future reference how, exactly would you determine the limit to be 1?
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