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anonymous
 5 years ago
test for convergence or divergence:
inf(sigma)n=1 [(5n3)/(n^22n+5)]
anonymous
 5 years ago
test for convergence or divergence: inf(sigma)n=1 [(5n3)/(n^22n+5)]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It converges. You can replace all of that with (for convergence tests only)\[\sum_{n=1}^{\infty}(1/(n^2)\] which converges according to the ptest. Convergent :D

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1It 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
 5 years ago
Best ResponseYou've already chosen the best response.0Why would you compare it to 1/n?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Because the degree difference between the top and the bottom is 1. So roughly the series behave like the harmonic series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good answer. Wasn't thinking :D
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