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

Use the Direct Comparison Test to determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}n}{{n}^{3}}$
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Obiajulu
Given:
The series, $\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}n}{{n}^{3}}$
To determine the convergence or divergence of the series using the Direct Comparison Test.
The Direct Comparison Test:
Let, ${x}_{n}\ge 0,{y}_{n}\ge 0$ for all $n\in N$ and ${x}_{n}\le {y}_{n}\mathrm{\forall }n$. Then (i) if $\sum {y}_{n}$ is convergent $⇒\sum {x}_{n}$ is convergent
(ii) If $\sum {x}_{n}$ is divergent $⇒\sum {y}_{n}$ is divergent
Let, $\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(n\right)}{{n}^{3}}$
Here, consider ${x}_{n}={\mathrm{sin}}^{2}n$ and ${y}_{n}=\frac{1}{{n}^{3}}$
$⇒\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(n\right)}{{n}^{3}}=\sum _{n=1}^{\mathrm{\infty }}|\frac{{\mathrm{sin}}^{2}\left(n\right)}{{n}^{3}}|$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{|{\mathrm{sin}}^{2}\left(n\right)|}{{n}^{3}}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{|\mathrm{sin}\left(n\right){|}^{2}}{{n}^{3}}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{3}}$
That is, $\frac{|\mathrm{sin}\left(n\right){|}^{2}}{{n}^{3}}\le \frac{1}{{n}^{3}}$
And by p-series test, the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{3}}$ is convergent.
Therefore, by the Direct Comparison test $\sum {x}_{n}$ is convergent.
Thus, by using the Direct Comparison test the series, $\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(n\right)}{{n}^{3}}$ is absolutely convergent
Hence, the given series is convergent.
Jeffrey Jordon