Exercises about series

Question

Using the Cauchy criterion,specify the nature of the series $$\sum_{k=1}^\infty \frac{k+1}{3k+2}=$$

Study the convergence(or divergence) of this series with positive terms $$\sum_{k=1}^\infty \frac{\sqrt[3]{k+1}-\sqrt[3]{k}}{k^2}=$$

Answer 1

We have that

$$ \frac{k+1}{3k+2} \to \frac13$$

therefore the series diverges and

$$\frac{\sqrt[3]{k+1}-\sqrt[3]{k}}{k^2}=\frac{\sqrt[3]{(k+1)^2}+\sqrt[3]{(k+1)k}+\sqrt[3]{k^2}}{k^2}=\frac{3k^\frac23}{k^2} \sim \frac{1}{ k^\frac43}$$

therefore the series converges by limit comparison test with $\sum \dfrac{1}{ k^\frac43}$.