Use the alternating series test to determine the convergence of the series sum_{n=1}^infty(-1)^nsin^2n

Use the alternating series test to determine the convergence of the series sum_{n=1}^infty(-1)^nsin^2n

Question
Series
asked 2020-11-02
Use the alternating series test to determine the convergence of the series
\(\sum_{n=1}^\infty(-1)^n\sin^2n\)

Answers (1)

2020-11-03
We have given a series,
\(\sum_{n=1}^\infty(-1)^n\sin^2n\)
We know that for the alternating series test, series should be given in forms,
\(\sum_{n=1}^\infty(-1)^na_n\quad\text{or}\quad\sum_{n=1}^\infty(-1)^{n+1}a_n\)
Our series is given in the form,
\(\sum_{n=1}^\infty(-1)^na_n\)
Conditions for alternate series are, \(a_n>0\)
And
\(\lim_{n=\infty}a_n=0\)
And series should be in decreasing order to be a series convergent.
\(\lim_{n=\infty}\sin^2n\cdot n\ne0\)
Hence, the given series is not convergent.
0

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