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Thread: Help with Convergence Test

  1. #1
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    Help with Convergence Test

    Help with Convergence Test-capture.png

    Trying to show that this is conditionally convergent. However, i know it is convergent, but i'm struggling to show that the absolute value of this series diverges.

    I am trying to use the Limit Comparison Test with 1/(n^1/2) or 1/n.

    The limits either go to 0 or infinity which is of no help.

    From my textbook are the conditions for the test:
    If L > 0, then  an converges if and only if  bn converges.
    If L = ∞ and  an converges, then  bn converges.
    If L = 0 and  bn converges, then  an converge
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  2. #2
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    Re: Help with Convergence Test

    For $n>10000$, you have $n>\sqrt{n}(\ln n)^2$. So,

    $$\sum_{n=10000}^\infty \dfrac{1}{\sqrt{n}(\ln n)^2} > \sum_{n=10000}^\infty \dfrac{1}{n} = \infty$$
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  3. #3
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    Re: Help with Convergence Test

    So basically plug a number and check it? Usually, my professor wants us to show inequality proof by simplifying the inequality more. I'm not sure if we can just plug numbers in. Is that the only way?
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    Re: Help with Convergence Test

    Quote Originally Posted by lc99 View Post
    So basically plug a number and check it? Usually, my professor wants us to show inequality proof by simplifying the inequality more. I'm not sure if we can just plug numbers in. Is that the only way?
    He didn't just plug in a number and check it.

    He established that there is a subseries of the original that diverges and thus the original must diverge as well.
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  5. #5
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    Re: Help with Convergence Test

    i understand, but this is the direct comparison test correct? On the exam, shouldn't we try to prove the inequality?
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    Re: Help with Convergence Test

    Quote Originally Posted by lc99 View Post
    So basically plug a number and check it? Usually, my professor wants us to show inequality proof by simplifying the inequality more. I'm not sure if we can just plug numbers in. Is that the only way?
    I suppose you could compute that the smallest $N$ such that $n>N \Rightarrow \dfrac{1}{\sqrt{n}\ln(n)^2} > \dfrac{1}{n}$ is $N=5504$

    It isn't true for values of $N$ less than this.

    But why bother. It doesn't matter that the divergent subseries starts at 5504, or 10000, as long as we know it exists.
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  7. #7
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    Re: Help with Convergence Test

    You want to show that $n>\sqrt{n}(\ln n)^2$.

    Find
    $$\begin{align*}\lim_{n\to \infty} \dfrac{\sqrt{n}(\ln n)^2}{n} & = \lim_{n\to \infty} \dfrac{2\ln n \tfrac{1}{n} }{\tfrac{1}{2\sqrt{n}}} \\ & = \lim_{n\to \infty} \dfrac{4\ln n}{\sqrt{n}} \\ & = \lim_{n\to \infty} \dfrac{\tfrac{4}{n}}{\tfrac{1}{2\sqrt{n}}} \\ & = \lim_{n\to \infty} \dfrac{8}{\sqrt{n}} = 0\end{align*}$$

    This shows that $n\gg \sqrt{n}(\ln n)^2$

    In fact, using this method, you can show that for any $r>0$ and any $k\in \mathbb{N} $ you have $n^r\gg (\ln n)^k $.
    Last edited by SlipEternal; Apr 8th 2018 at 08:40 PM.
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