# How do you test this series? sum for n=1 to infinity sin^2(1/n) convergence , by using limit comparison test with cn=1/n^2 .

This series is convergent, because

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To test the convergence of the series ( \sum_{n=1}^\infty \sin^2\left(\frac{1}{n}\right) ) using the limit comparison test with ( c_n = \frac{1}{n^2} ):

- Calculate the limit as ( n ) approaches infinity of ( \frac{\sin^2\left(\frac{1}{n}\right)}{\frac{1}{n^2}} ).
- If the limit is a finite positive number, then the series ( \sum_{n=1}^\infty \sin^2\left(\frac{1}{n}\right) ) converges or diverges in the same manner as the series ( \sum_{n=1}^\infty \frac{1}{n^2} ).
- Since ( \sum_{n=1}^\infty \frac{1}{n^2} ) is a convergent ( p )-series with ( p = 2 ), if the limit from step 1 is finite and positive, then the original series also converges.

The limit comparison test is based on the fact that if ( \lim_{n \to \infty} \frac{a_n}{b_n} = L ), where ( L ) is a finite positive number, then the series ( \sum a_n ) and ( \sum b_n ) either both converge or both diverge.

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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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