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.

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.