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Math Help - Proving convergence and divergence by Comparison Test

  1. #1
    Junior Member EliteAndoy's Avatar
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    Proving convergence and divergence by Comparison Test

    Hi Everyone! Today we were asked to prove \sum_{n=1}^{\infty}\frac{\ln(n)}{n^2} by comparison test. This is a pretty easy check, since the function is positive, decreasing, and continuous at interval (1,\infty). Integral test tells us that this integral shall converge. But the catch is, what would be the sequence input that shall be compared to the original sequence input so that it can be proven by comparison test. I mean, I can't say that \frac{\ln(n)}{n^2} behaves like \frac{n}{n^2} since n behaves way faster than \ln(n). This comparison also makes the comparison test useless since \frac{n}{n^2}\geq \frac{ln(n)}{n^2}. On the other hand, you can't also say that \frac{\ln(n)}{n^2} behaves like \frac{1}{n^2} since \ln(n) is a bit faster than 1. So what I was thinking is saying that \frac{\ln(n)}{n^2} behaves like \frac{verylargenumber}{n^2} as n \to \infty. The thing is, however, making the numerator a constant will make the series convergent, but in the long run, will never approach \infty. So anyone got any idea on what sequence input should I compare the original sequence to, so that I can prove it by Comparison Test. Thanks everyone in advance!
    Last edited by EliteAndoy; March 29th 2013 at 05:20 PM. Reason: Typo
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    Re: Proving convergence and divergence by Comparison Test

    Why use the comparison test when you already have proven its convergence using the integral test?
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    Junior Member EliteAndoy's Avatar
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    Re: Proving convergence and divergence by Comparison Test

    We were asked to do so. I was suspecting that there is some crazy stuff that I should do to prove it using comparison test, and I won't be able to sleep tonight unless I prove this using comparison test.
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    Re: Proving convergence and divergence by Comparison Test

    You could say \displaystyle \sum{\frac{\ln{(n)}}{n^2}} < \sum{\frac{\sqrt{n}}{n^2}} since \ln{(n)} < \sqrt{n} for all n > 0. Simplifying the RHS gives \displaystyle \sum{ \frac{1}{n^{\frac{3}{2}}} }, which is a convergent p series.
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    Re: Proving convergence and divergence by Comparison Test

    Quote Originally Posted by EliteAndoy View Post
    Hi Everyone! Today we were asked to prove \sum_{n=1}^{\infty}\frac{\ln(n)}{n^2} by comparison test.

    \frac{\ln(n)}{n^2}=\frac{2\ln(\sqrt{n})}{n^2}\le \frac{2\sqrt{n}}{n^2}\le\frac{2}{n^{3/2}}
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    Junior Member EliteAndoy's Avatar
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    Re: Proving convergence and divergence by Comparison Test

    Woah, never thought of setting ln(n) as 2ln(n^(1/2)). Both of your solutions really worked, but Plato cleared it up for me once and for all. Thank you guys for letting me sleep tonight!
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