# How do you show whether #sum_(n=2)^oo 1/ln^3(n)# converges or diverges?

Now, looking at the condensed sum, we have

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Reapplying L'Hospitals (we can see a pattern forming):

L'Hospital's once more:

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To determine whether the series (\sum_{n=2}^{\infty} \frac{1}{\ln^3(n)}) converges or diverges, we can use the integral test.

Let (f(x) = \frac{1}{\ln^3(x)}).

We integrate (f(x)) from (x = 2) to (x = \infty), which gives us: [\int_{2}^{\infty} \frac{1}{\ln^3(x)} dx]

To evaluate this improper integral, we can use substitution. Let (u = \ln(x)), then (du = \frac{1}{x} dx). Substituting these values, the integral becomes: [\int \frac{1}{u^3} du]

Integrating, we get: [\lim_{b \to \infty} \left[-\frac{1}{2u^2}\right]_{\ln(2)}^{\ln(b)}]

Evaluating the limit, we have: [\lim_{b \to \infty} \left[-\frac{1}{2\ln^2(b)} + \frac{1}{2\ln^2(2)}\right]]

Since (\lim_{b \to \infty} \frac{1}{2\ln^2(b)} = 0), the integral converges if and only if (\frac{1}{2\ln^2(2)}), a constant, converges. Since the constant is finite, the integral converges, and by the integral test, the series (\sum_{n=2}^{\infty} \frac{1}{\ln^3(n)}) converges.

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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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