Answer
Divergent
Work Step by Step
Since $$\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\frac{1}{17}+\cdots=\sum_{n=1}^{\infty} \frac{1}{3 n+2}$$
The function $f(x)=\dfrac{1}{3 x+2}$ is contimuous, positive, and decreasing on $[1,\infty)$, so by using Integral Test, we get
\begin{align*}
\int_{1}^{\infty} \frac{1}{3 x+2} d x&=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{3 x+2} d x\\
&=\lim _{t \rightarrow \infty}\left(\frac{1}{3} \ln |3 x+2|\right)\bigg|_{1}^{t}\\
&=\frac{1}{3} \lim _{t \rightarrow \infty}(\ln (3 t+2)-\ln 5)\\
&=\infty
\end{align*}
Hence $\displaystyle\sum_{n=1}^{\infty} \frac{1}{3 n+2}$ diverges