Chapter 9 - Infinite Series - 9.6 Alternating Series; Absolute And Conditional Convergence - Exercises Set 9.6 - Page 646: 3

Divergent

We have: $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{k+1}{3k+1}\\=\lim\limits_{k \to \infty} \dfrac{1+\dfrac{1}{k}}{3+\dfrac{1}{k}}\\= \dfrac{1+\dfrac{1}{\infty}}{3+\dfrac{1}{\infty}}\\=\dfrac{1}{3} \ne 0$ We conclude that the given series is divergent by the divergence test.