# How do you determine if the series the converges conditionally, absolutely or diverges given #sum_(n=1)^oo (cos(npi))/n^2#?

See below.

The series

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To determine if the series ( \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} ) converges conditionally, absolutely, or diverges, we need to analyze the convergence of both ( \sum_{n=1}^\infty |\frac{\cos(n\pi)}{n^2}| ) and ( \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} ).

First, consider ( \sum_{n=1}^\infty |\frac{\cos(n\pi)}{n^2}| ). Since ( |\cos(n\pi)| = 1 ) for all ( n ) (as ( \cos(n\pi) ) alternates between ( -1 ) and ( 1 )), this series reduces to ( \sum_{n=1}^\infty \frac{1}{n^2} ), which is a convergent ( p )-series with ( p = 2 ). Therefore, ( \sum_{n=1}^\infty |\frac{\cos(n\pi)}{n^2}| ) converges absolutely.

Next, we need to examine ( \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} ). Since ( \cos(n\pi) ) alternates between ( -1 ) and ( 1 ) for different values of ( n ), the series ( \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} ) is conditionally convergent. This is because the alternating sign causes the terms to alternate in sign, which prevents the series from converging absolutely, but the magnitude of the terms decreases monotonically to zero, satisfying the conditions for convergence by the Alternating Series Test.

In summary:

- ( \sum_{n=1}^\infty |\frac{\cos(n\pi)}{n^2}| ) converges absolutely.
- ( \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} ) converges conditionally.

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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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