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tells convergence or divergence of an improper integral

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tells convergence or divergence of an improper integral

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x^n: if n is greater than or equal to 1, the function converges to a constant value.
x^n: if n is less than 1, the function diverges.

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Change series into a f(x) function and take the integral like an improper integral.

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The functions must be:
continuous
positive
Decreasing

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0 < An < Bn
If the series Bn converges then series An converges
If the series An diverges then series Bn diverges

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limit as n goes to infinity (a[n]/b[n])

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Lim as n goes to infinity An / Bn = L
If the series An converges then series Bn converges
If sereis An diverges then series Bn diverges
L=Positive finite value

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Check if b(n) is decreasing and whether the limit of b(n) goes to zero. If both conditions are met, the series converges

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lim n goes to infinity                    | An + 1/ An|

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lim n goes to infinity                    | An + 1/ An|

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converges

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test with absolute values to tell if converge or diverge.

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If ratio test diverges, means conditional. Also need to test the alternate series test to see if the series is conditionally convergent. If the ratio test gives convergence, then absolute convergence.

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diverges

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inconclusive

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Based on Bn's behavior, Bn's behavior would be the same as An if there is a positive finite result from the limit.

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0<Bn<An<known divergent<given integral