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

Answer 1

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.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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