# How do you use the limit comparison test for #sum( n^3 / (n^4-1) ) # from n=2 to #n=oo#?

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To use the limit comparison test for the series ( \sum_{n=2}^{\infty} \frac{n^3}{n^4 - 1} ):

- Identify a simpler series that you know converges or diverges.
- Take the limit of the ratio of the given series to the simpler series as ( n ) approaches infinity.
- If the limit is a positive finite number, then both series either converge or diverge together. If the limit is zero or infinite, the comparison test is inconclusive.

Let's choose the simpler series ( \sum_{n=2}^{\infty} \frac{1}{n} ) as our comparison series.

Now, take the limit: [ \lim_{n \to \infty} \frac{\frac{n^3}{n^4 - 1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^4}{n^4 - 1} ]

[ = \lim_{n \to \infty} \frac{1}{1 - \frac{1}{n^4}} = 1 ]

Since the limit is a positive finite number, we conclude that both series either converge or diverge together. Since ( \sum_{n=2}^{\infty} \frac{1}{n} ) is a harmonic series which diverges, the original series ( \sum_{n=2}^{\infty} \frac{n^3}{n^4 - 1} ) also diverges by the limit comparison test.

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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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