Convergence of $\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}$

Question

To show that $$\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}$$ is convergent My attempt: Expanding the first term using binomial series and simplifying, I tried comparison test with $b_{n}=\frac{1}{n^2}$ $$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\log n}\right)n^2 $$ $$=\lim_{n \to \infty}\frac{n\left(\frac13\frac{1 }{n^{3}}-\frac19\frac{1 }{n^{6}}+\dots\right)n^2}{\log n}$$ I could simplify this using binomial theorem but I don't know how to deal with the $\log$ term in the denominator

Answer 1

$$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\ln n}\right)n^2=\lim_{n\rightarrow+\infty}\frac{n^2}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2\right)\ln{n}}=0,$$ which says that it converges because $$\sum_{k=1}^{+\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$$ Also, you can write $$\sum_{k=3}^{+\infty}\frac{(n^3+1)^\frac13-n}{\ln n}<\sum_{k=3}^{\infty}\frac{1}{3k^2}.$$