Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges 
Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.

I showed it using Abel's theorem and limit comparison test. Any other simpler method?

$$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\ln n}\right)n^2=\lim_{n\rightarrow+\infty}\frac{n^2}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2\right)\ln{n}}=0,$$
then for $n$ large enough we have 
$$\frac{(n^3+1)^\frac13-n}{\ln n}\le \frac{1}{ n^2}$$
the result follows by comparison test. which says that it converges because $$\sum_{k=1}^{+\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$$

