anonymous
  • anonymous
test for convergence or divergence: inf(sigma)n=1 [(5n-3)/(n^2-2n+5)]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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
watchmath
  • watchmath
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.
anonymous
  • anonymous
Why would you compare it to 1/n?

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watchmath
  • watchmath
Because the degree difference between the top and the bottom is 1. So roughly the series behave like the harmonic series.
anonymous
  • anonymous
Good answer. Wasn't thinking :D

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