# How do I use the Limit Comparison Test on the series #sum_(n=1)^oosin(1/n)# ?

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To use the Limit Comparison Test on the series (\sum_{n=1}^\infty \sin\left(\frac{1}{n}\right)), you need to compare it to a known series whose convergence behavior is already established. Choose a series (b_n) such that:

- (b_n > 0) for all (n).
- (\lim_{n \to \infty} \frac{a_n}{b_n}) exists and is finite, where (a_n = \sin\left(\frac{1}{n}\right)).

If (\lim_{n \to \infty} \frac{a_n}{b_n}) is a nonzero finite number, then both series either converge or diverge.

Typically, for the series (\sum_{n=1}^\infty \sin\left(\frac{1}{n}\right)), a common choice for comparison is (\sum_{n=1}^\infty \frac{1}{n}) (the harmonic series), as their behavior is related.

Apply the Limit Comparison Test:

- Let (b_n = \frac{1}{n}).
- Calculate the limit (\lim_{n \to \infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}).
- If the limit is finite and nonzero, the series (\sum_{n=1}^\infty \sin\left(\frac{1}{n}\right)) behaves the same as (\sum_{n=1}^\infty \frac{1}{n}).

You can proceed with the calculation to determine the convergence behavior of the series.

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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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- Find a series expansion for? : # (1-3x)^(2/3) #
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