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

1 Answer

Jan 27, 2017

We have that

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

as both the series at the second member are absolutely convergent, then also our series is convergent.

Explanation:

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