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Math Help - infinite series

  1. #1
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    infinite series

    test for convergence/divergence of the series \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}

    i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by buttonbear View Post
    test for convergence/divergence of the series \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}

    i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)

    Direct comp, since e^{1/n}\to 1 as n\to\infty

    use 0<e^{1/n}<100 so \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2} < 100\sum_{n =1}^{\infty}\frac{1}{n^2}<\infty

    where the 100 is larger than what you need, but it get the point across.
    The max of e^{1/n} is actually e. So, any number e or larger works, but that's not important.
    Last edited by matheagle; March 28th 2009 at 06:56 PM.
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  3. #3
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    Quote Originally Posted by buttonbear View Post
    test for convergence/divergence of the series \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}

    i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)
    To answer your question, yes the integral test can be used and also a limit comparision with \sum_{n=1}^\infty \frac{1}{n^2}. It might be a bit easier the a direct comparision test.
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  4. #4
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    okay, so, just to be clear..it CAN be compared to the p-series? i'm just not 100% sure that i understand all of that..and i need to be able to explain my work. thanks again!
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  5. #5
    MHF Contributor matheagle's Avatar
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    Yes, I did a direct comp test to the p=2 series.
    You can do a limit comp test here too, with p=2 also.
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  6. #6
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    thanks so much
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