How do you test the series #Sigma 1/sqrt(n^3+4)# from n is #[0,oo)# for convergence?

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Answer 1 · 2017-05-06T16:20:11

The series:

#sum_(n=0)^oo 1/sqrt(n^3+4)#

is convergent.

The series:

#sum_(n=0)^oo 1/sqrt(n^3+4)#

has positive terms. Now consider that if we lower the denominator the quotient increases, so:

#0 < 1/sqrt(n^3+4) < 1/sqrt(n^3)#

We know however that the series:

#sum_(n=1)^oo 1/sqrt(n^3) = sum_(n=1)^oo 1/n^(3/2)#

is convergent based on the p-series test, so also our series is convergent based on direct comparison.