Answer
Converges
Work Step by Step
Given
$$\sum_{k=0}^{\infty}\left(\frac{k}{3 k+1}\right)^{k}$$
By using the Root Test, we get:
\begin{align*}
\rho&=\lim _{k \rightarrow \infty}\sqrt[k]{a_{k}}\\
&=\lim _{k\rightarrow \infty}\sqrt[k]{ \left(\frac{k}{3 k+1}\right)^{k}}\\
&= \lim _{k \rightarrow \infty}\frac{k}{3 k+1}\\
&=\frac{1}{3}<1
\end{align*}
Thus the series converges.
