# Does the series $\sum_{n=1}^\infty \left(\frac{1}{\sqrt[3]{n}}-\sqrt[3]{\ln(\frac{n+1}{n})}\right)$ diverge or converge?

Sum: $$\sum_{n=1}^\infty \left(\frac{1}{\sqrt[3]{n}}-\sqrt[3]{\ln(\frac{n+1}{n})}\right)$$ tried to simplify: $$\sqrt[3]{\ln(\frac{n+1}{n})} \sim \frac{1}{\sqrt[3]{n}}$$ and I got zero here: $$\lim_{n\to \infty} \frac{1}{\sqrt[3]{n}} - \frac{1}{\sqrt[3]{n}} = 0$$

• By the comparison test, the series diverges! – Jan Jun 16 '17 at 10:40
• is it $+$ in the series or $-$? – Julián Aguirre Jun 16 '17 at 10:47

I assume that there is a typo and that the series is with a $-$ instead of a $+$. For $x$ near $0$ we have $$\sqrt[3]{\ln(1+x)}=\sqrt[3]{x-\frac{x^2}{2}+O(x^3)}=\sqrt[3]{x}\Bigl(1-\frac{x}{6}+O(x^2))\Bigr).$$ Then, with $x=1/n$ we get $$\frac{1}{\sqrt[3]{n}}-\sqrt[3]{\ln(\frac{n+1}{n})}=\frac{1}{\sqrt[3]{n}}-\frac{1}{\sqrt[3]{n}}\Bigr(1-\frac{1}{6\,n}+O(\frac{1}{n^2})\Bigr)=\frac{1}{6\,n^{4/3}}+O(\frac{1}{n^{7/3}}).$$ Since $4/3>1$, the series converges.

• how did you remove square root here? $$\sqrt[3]{x-\frac{x^2}{2}+O(x^3)}=\sqrt[3]{x}\Bigl(1-\frac{x}{6}+O(x^2))\Bigr)$$ – Дмитро Пінтак Jun 16 '17 at 15:19
• The Taylor series of $\sqrt[3]{1+x}$. – Julián Aguirre Jun 18 '17 at 12:07

One has $\sum_{n=1}^m \frac{1}{\sqrt[3]{n}}+\sqrt[3]{\ln(\frac{n+1}{n})} \geq \sum_{n=1}^m \frac{1}{\sqrt[3]{n}} \geq \sum_{n=1}^m \frac{1}{{n}} \rightarrow \infty$, as $m$ approaches $\infty$.

So the series diverges.

(Obviously this answer is obsolete now that OP has edited his question. I'm still going to leave this here just in case somebody is interested in the divergence of the series with "+" instead of "-" :) )

$$\sum_{n=1}^\infty (\frac{1}{\sqrt[3]{n}}+\sqrt[3]{ln(\frac{n+1}{n})})\geq\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n}},$$ since the the inequality holds for all terms in the series. And $$\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n}}$$ diverges by: comaparison and Riemann's rules $\frac{1}{3}<1$

• if you had $>$ instead of $<$, it would have been correct and accepted. – farruhota Jun 16 '17 at 11:09