# Using the integral test, how do you show whether #sum ln(n)/(n)^2# diverges or converges from n=1 to infinity?

See the explanation.

I assume that the difficulty is in integrating

So we get

Or perhaps the difficulty is in evaluating

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To determine the convergence or divergence of the series (\sum \frac{\ln(n)}{n^2}) from (n = 1) to (\infty) using the integral test, we compare it with the integral of the corresponding function.

First, note that (\frac{\ln(n)}{n^2}) is positive and decreasing for all (n \geq 1). Thus, we can apply the integral test.

Let (f(x) = \frac{\ln(x)}{x^2}). To find the integral, we evaluate (\int_1^\infty \frac{\ln(x)}{x^2} , dx).

Using integration by parts with (u = \ln(x)) and (dv = \frac{1}{x^2} , dx), we have:

[ \int \frac{\ln(x)}{x^2} , dx = -\frac{\ln(x)}{x} + \int \frac{1}{x^2} , dx = -\frac{\ln(x)}{x} + \frac{1}{x} + C]

Evaluating the definite integral from (1) to (\infty), we get:

[ \int_1^\infty \frac{\ln(x)}{x^2} , dx = \lim_{b \to \infty} \left(-\frac{\ln(b)}{b} + \frac{1}{b} - \left(-\frac{\ln(1)}{1} + \frac{1}{1}\right)\right)]

[= \lim_{b \to \infty} \left(-\frac{\ln(b)}{b} + \frac{1}{b} + 1\right) = 1 ]

Since the integral converges to a finite value, by the integral test, the series (\sum \frac{\ln(n)}{n^2}) also 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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