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Thread: Converging and Diverging Infinite Series help

  1. #1
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    Converging and Diverging Infinite Series help

    Determine whether the series converges or diverges:

    1. $\displaystyle \sum_{n=2} \frac{ln(n)}{n}$


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    I tried the $\displaystyle n$-th term test:

    $\displaystyle \lim_{n \to \infty} \frac{ln(n)}{n}$
    $\displaystyle \lim_{n \to \infty} (\frac{1}{n})(ln(n))$

    As the $\displaystyle n$ in $\displaystyle \frac{1}{n}$ approaches $\displaystyle \infty$, the denominator gets larger and larger, therefore making the value of $\displaystyle \frac{1}{n}$ smaller and smaller. So then you're essentially multiplying by zero.

    So I need to try a new test, but what test do I try?

    ------------------------------

    2. $\displaystyle \sum_{n=1} \frac{1}{(ln2)^n}$

    Can someone give me a hint to help me get started on this one?
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  2. #2
    Member Ruun's Avatar
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    Hi!

    Check Convergence tests - Wikipedia, the free encyclopedia for other convergence tests you can try.

    The second one, remember that $\displaystyle ln(2)$ it's a number. Then

    $\displaystyle \sum_{n=1} \frac{1}{(ln2)^n} = \sum_{n=2} \frac{1^n}{ln(2)^n}$

    Maybe the second term will make think you in a very special case of series
    .
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Cursed View Post
    Determine whether the series converges or diverges:

    1. $\displaystyle \sum_{n=2} \frac{ln(n)}{n}$

    as $\displaystyle \ln(n)$ is positive and strictly increasing we have $\displaystyle \frac{\ln(n)}{n}>\frac{\ln(2)}{n}$ and as the harmonic series is divergent your series is divergent.

    CB
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