Question

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

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asked 2021-01-13
Use the Direct Comparison Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{\sin^2n}{n^3}\)

Answers (1)

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.
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