Test the series for convergence or divergence.
{eq}\displaystyle \sum_{n\ =\ 1}^{\infty} (-1)^n \sin \left (\dfrac \pi n\right) {/eq}
Question:
Test the series for convergence or divergence.
{eq}\displaystyle \sum_{n\ =\ 1}^{\infty} (-1)^n \sin \left (\dfrac \pi n\right) {/eq}
Alternating Series Test:
Alternating series test is very useful in checking convergence of conditionally convergent series i.e. those convergent series which are not absolutely convergent. This test goes as follows:
Suppose {eq}\left\{ a_n \right\} {/eq} is a monotonic decreasing sequence of positive real numbers and {eq}\lim\limits_{n \to \infty}a_n=0 {/eq}, then {eq}\sum\limits_{n=1}^{\infty}\left( -1 \right)^n a_n {/eq} is convergent.
Answer and Explanation: 1
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View this answerThe given series {eq}\displaystyle \sum\limits_{n=1}^{\infty}(-1)^n \sin \left (\dfrac \pi n\right). {/eq}
Let, {eq}\displaystyle a_n=\sin \left...
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Chapter 21 / Lesson 5Learn the convergence and divergence tests for an infinite series. See how to use comparison tests to determine if a series is convergent or divergent with examples.