 # Use the alternating series test to determine the convergence of the series sum_{n=1}^infty(-1)^nsin^2n Kyran Hudson 2020-11-02 Answered
Use the alternating series test to determine the convergence of the series
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{\mathrm{sin}}^{2}n$
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We have given a series,
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{\mathrm{sin}}^{2}n$
We know that for the alternating series test, series should be given in forms,
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{a}_{n}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}{a}_{n}$
Our series is given in the form,
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{a}_{n}$
Conditions for alternate series are, ${a}_{n}>0$
And
$\underset{n=\mathrm{\infty }}{lim}{a}_{n}=0$
And series should be in decreasing order to be a series convergent.
$\underset{n=\mathrm{\infty }}{lim}{\mathrm{sin}}^{2}n\cdot n\ne 0$
Hence, the given series is not convergent.
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