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anonymous

  • 5 years ago

test for convergence or divergence: inf(sigma)n=1 [(5n-3)/(n^2-2n+5)]

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  1. anonymous
    • 5 years ago
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    It converges. You can replace all of that with (for convergence tests only)\[\sum_{n=1}^{\infty}(1/(n^2)\] which converges according to the p-test. Convergent :D

  2. watchmath
    • 5 years ago
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    It is divergent. Use the limit comparison test! compare to \(b_n=\frac{1}{n}\). The limit of \(a_n/b_n\) is 5 but \(\sigma b_n\) is divergent. So the original is divergent as well.

  3. anonymous
    • 5 years ago
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    Why would you compare it to 1/n?

  4. watchmath
    • 5 years ago
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    Because the degree difference between the top and the bottom is 1. So roughly the series behave like the harmonic series.

  5. anonymous
    • 5 years ago
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    Good answer. Wasn't thinking :D

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