Answer
Convergent
Work Step by Step
Given
$$\sum_{n=1}^{\infty} \frac{1}{n^{3}+8}$$
Use the Limit Comparison Test with $a_n =\dfrac{1}{n^{3}+8}$ and $b_n=\dfrac{ 1}{ n^{3}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^3}{n^{3}+8}\\
&=\lim _{n \rightarrow \infty}\frac{n^3/n^3}{n^{3}/n^3+8/n^3}\\
&=\lim _{n \rightarrow \infty}\frac{1}{1+8/n^3}\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{ 1}{ n^{3}}$ is convergent ($p$ -series, $p>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3}+8}$ is also convergent.
