Answer
Diverges
Work Step by Step
The kth term formula of the series is given by: $a_k=\dfrac{k^2+2k+1}{3k^2+1}$
Thus, $\lim\limits_{k \to \infty}a_k=\lim\limits_{k \to \infty}\dfrac{k^2+2k+1}{3k^2+1}\\=\lim\limits_{k \to \infty}\dfrac{1+2/k+1/k^2}{3+1/k^2}\\=\dfrac{1}{3}$
Therefore, we can see that the limit of kth term is not zero, this implies that the series diverges.
