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Math Help - Finding convergence/divergence

  1. #1
    Junior Member
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    Finding convergence/divergence

    Hei.

    So I'm stuck on this series problem where I have:

    1 / (n + sqr.root (n))

    Starting from n = 1.

    Can I simply conclude here that since 1/n is the lowest fraction which will give a divergent series the abovementioned series has to converge? (according to the p-test?).

    I just can't find a way to actually calculate this. I tried to treat it as a telescoping series with 1/(sqr.root (x)*(sqr.root(x) + 1). However, this did not lead to any conclusive answer.

    Any help will be greatly appreciated!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    the series diverges because

    \frac{1}{n+\sqrt{n}}>\frac{1}{n+n}=\frac1{2n}.
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  3. #3
    Junior Member
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    Thanks a lot! Why do I always miss those easy ways out?
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