Answer
$0.83253$
$R_{10}=0.005$
Work Step by Step
Given
$$\sum_{n=1}^{\infty} \frac{\sin ^{2} n}{n^{3}}$$
The first $10$ terms are:
\begin{align*}
\sum_{n=1}^{10} \frac{\sin ^{2} n}{n^{3}}&=\frac{\sin ^{2} 1}{1}+\frac{\sin ^{2} 2}{8}+\frac{\sin ^{2} 3}{27}+\cdots+\frac{\sin ^{2} 10}{1000}\\
&\approx 0.83253
\end{align*}
We estimate the error:
$$ \frac{\sin ^{2} n}{n^{3}}\leq\frac{1}{ n^{3 }} $$
Then
\begin{align*}
R_{10} \leq T_{10} &\leq \int_{10}^{\infty} \frac{1}{x^{3}} d x\\
&=\lim _{t \rightarrow \infty}\frac{-1}{2x^2}\bigg|_{10}^{t}\\
&= \lim _{t \rightarrow \infty}\left(-\frac{1}{2t^2}+\frac{1}{200}\right)\\
&=\frac{1}{200}=0.005
\end{align*}
