# How do you use the limit comparison test to determine if #Sigma 1/(n(n^2+1))# from #[1,oo)# is convergent or divergent?

The series:

is convergent.

The limit comparison test states that if we have two series with positive terms:

where the limit:

exists and is finite, then if one series converges, the other is also convergent.

Given:

we can choose:

that we know is convergent based on the p-series test and verify that:

So the series:

is convergent.

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To use the limit comparison test to determine the convergence or divergence of the series ( \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)} ), we compare it to a known series whose convergence behavior is understood.

Let's compare it to the series ( \sum_{n=1}^{\infty} \frac{1}{n^3} ).

We calculate the limit:

[ \lim_{n \to \infty} \frac{\frac{1}{n(n^2+1)}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{n^3}{n(n^2+1)} = \lim_{n \to \infty} \frac{n^2}{n^2+1} ]

This limit is equal to 1, which means that the two series behave similarly. Since the series ( \sum_{n=1}^{\infty} \frac{1}{n^3} ) is known to be convergent (it is a p-series with ( p = 3 ) and ( p > 1 )), by the limit comparison test, the original series ( \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)} ) is also convergent.

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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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