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?
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?
Write $$\frac{(n^3+1)^{\frac{1}{3}}-n}{\log n} = \frac{n^3+1-n^3}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)} = \frac{1}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)}$$ and observe that the denominator goes at $0$ faster than $\frac{1}{n^2}.$
Note that
$$(n^3+1)^\frac13=n\left(1+\frac1{n^3}\right)^\frac13\sim n+\frac1{3n^2}$$
thus
$$\frac{(n^3+1)^\frac13-n}{\log n}\sim \frac{1}{3n^2\log n}$$
and thus since $\sum\limits_{n=1}^{\infty} \frac{1}{3n^2\log n}$ converges also $\sum\limits_{n=1}^{\infty} \frac{(n^3+1)^\frac13-n}{\log n}$ converges by comparison with it.
$$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\ln n}\right)n^2=\lim_{n\rightarrow+\infty}\frac{n^2}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2\right)\ln{n}}=0,$$
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 $$\sum_{k=1}^{+\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$$