## Calculus: Early Transcendentals 8th Edition

The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Using the Comparison Test, the series is less than $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}}$ because the denominator of the original series is always slightly larger than the series on the right. Therefore, $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}}$ Re-arranging the exponents: $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{k^{1/3}}{k^{3/2}}$ $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{1}{k^{1/6}}$ We find that the series on the right side is a p-series with $p \gt 1$, so the series on the right side converges and therefore, the series on the left converges.