# Question # Use the Direct Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{sin^2n}{n^3}

Series
ANSWERED Use the Direct Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{\sin^2n}{n^3}$$ 2021-01-14
Given:
The series, $$\sum_{n=1}^\infty\frac{\sin^2n}{n^3}$$
To determine the convergence or divergence of the series using the Direct Comparison Test.
The Direct Comparison Test:
Let, $$x_n\geq0,y_n\geq0$$ for all $$n\in N$$ and $$x_n\leq y_n\forall n$$. Then (i) if $$\sum y_n$$ is convergent $$\Rightarrow\sum x_n$$ is convergent
(ii) If $$\sum x_n$$ is divergent $$\Rightarrow\sum y_n$$ is divergent
Let, $$\sum_{n=1}^\infty\frac{\sin^2(n)}{n^3}$$
Here, consider $$x_n=\sin^2n$$ and $$y_n=\frac{1}{n^3}$$
$$\Rightarrow\sum_{n=1}^\infty\frac{\sin^2(n)}{n^3}=\sum_{n=1}^\infty|\frac{\sin^2(n)}{n^3}|$$
$$=\sum_{n=1}^\infty\frac{|\sin^2(n)|}{n^3}$$
$$=\sum_{n=1}^\infty\frac{|\sin(n)|^2}{n^3}$$
$$=\sum_{n=1}^\infty\frac{1}{n^3}$$
That is, $$\frac{|\sin(n)|^2}{n^3}\leq\frac{1}{n^3}$$
And by p-series test, the series $$\sum_{n=1}^\infty\frac{1}{n^3}$$ is convergent.
Therefore, by the Direct Comparison test $$\sum x_n$$ is convergent.
Thus, by using the Direct Comparison test the series, $$\sum_{n=1}^\infty\frac{\sin^2(n)}{n^3}$$ is absolutely convergent
Hence, the given series is convergent.