How do you determine if #sum n^3/((n^4)-1)# from n=2 to #n=oo# is convergent?

Answer 1

It is divergent.

Since #n^3/(n^4-1)~n^3/n^4~1/n# that is the harmonic series, that diverges.
(#~# means asymptotic).
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Answer 2

You can determine if the series ∑(n^3/(n^4 - 1)) from n = 2 to ∞ is convergent by using the limit comparison test. First, find the limit as n approaches infinity of the term a_n = n^3 / (n^4 - 1). Then, choose a known convergent series ∑b_n that has positive terms and behaves similarly to a_n as n approaches infinity. If the limit of the ratio a_n / b_n exists and is a finite positive number, then both series converge or diverge together. If the limit is zero, the series ∑a_n converges. If the limit is infinity, ∑a_n diverges.

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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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