# How do you test for convergence for #sum ln(n)/n^2# for n=1 to infinity?

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You can test for convergence of the series ( \sum \frac{\ln(n)}{n^2} ) using the integral test:

- Integrate the function ( f(x) = \frac{\ln(x)}{x^2} ) from 1 to infinity.
- If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Performing the integral: [ \int_{1}^{\infty} \frac{\ln(x)}{x^2} , dx ]

Integration by parts: [ u = \ln(x) ] [ du = \frac{1}{x} , dx ] [ dv = \frac{1}{x^2} , dx ] [ v = -\frac{1}{x} ]

Applying integration by parts formula: [ \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 ]

Evaluate the integral from 1 to infinity: [ = \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} - \left(0 - 1\right) \right) ]

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

As ( b ) approaches infinity, ( \frac{\ln(b)}{b} ) approaches 0 (since the logarithmic function grows slower than any power of ( b )), and ( \frac{1}{b} ) approaches 0. Therefore, the integral converges to 1.

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

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To test for convergence of the series ( \sum \frac{\ln(n)}{n^2} ) from ( n = 1 ) to infinity, you can use the Integral Test. This test states that if ( f(x) ) is continuous, positive, and decreasing for all ( x \geq 1 ), then the series ( \sum_{n=1}^{\infty} f(n) ) converges if and only if the improper integral ( \int_{1}^{\infty} f(x) , dx ) converges.

In this case, ( f(x) = \frac{\ln(x)}{x^2} ). To apply the Integral Test, we integrate ( f(x) ) from 1 to infinity:

[ \int_{1}^{\infty} \frac{\ln(x)}{x^2} , dx ]

This integral can be evaluated using integration by parts or other methods. After integration, if the result is a finite value, then the series ( \sum \frac{\ln(n)}{n^2} ) converges. If the integral diverges, then the series also diverges.

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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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