Study the convergence of the series $\sum_{n=1}^\infty\left(\ln(n\tan\frac{1}{n})\right)^{1/3}$

Study the convergence of the $$\sum_{n=1}^\infty \ln^{1/3}\left(n\tan\frac{1}{n}\right)$$

I'm trying to study the convergence of the series and I need a little help. This is what I tried to do. I used the limit comparison test $$\lim_{n\to \infty}\frac{\ln(n\tan\frac{1}{n})}{n\tan\frac{1}{n}-1}=1$$ This implies that $$\sum_{n=1}^\infty\left(\ln(n\tan\frac{1}{n})\right)^\frac{1}{3} \sim \sum_{n=1}^\infty\left(n\tan\frac{1}{n}-1\right)^\frac{1}{3}$$

And this is where I got stuck because I don't know how to study the convergence of the new one. Should I try another method or should I proceed with this one? If so how can I do that?

The answer in the solution says that it diverges

• You are almost there: it happens that $f(x)=\frac1x\tan x-1$ is of order $x^2$ when $x\to0$, hence the $n$th term of the series is of order $\left(\frac1{n^2}\right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)\sim\frac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $\tan$ at order $3$, then your proof is complete. – Did Feb 1 at 7:16

For $$n$$ sufficiently large we have $$\tan {1\over n}={1\over n}+{1\over 3n^3}+\cdots$$ according to MaclaurinSeries therefore $$\forall n\in \Bbb N\quad,\quad \tan {1\over n}>{1\over n}+{1\over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$\left(n\tan {1\over n}-1\right)^{1\over 3}{>\left(1+{1\over 3n^2}-1\right)^{1\over 3}\\={1\over \sqrt[3]3 n^{2\over 3}}\\>{1\over 2n}}$$as $$\sum {1\over 2n}$$ diverges, so do $$\sum \left(n\tan {1\over n}-1\right)^{1\over 3}$$and $$\sum_{n=1}^\infty \ln^{1/3}\left(n\tan\frac{1}{n}\right)$$
• I understand what you tried to do but I don't think you that's the Maclaurin series for $\tan\frac{1}{n}$ shouldn't it be $\tan\frac{1}{n}=\frac{1}{n}+\frac{1}{3n^3}+\frac{2}{15n^5}+...$ – J.Dane Feb 2 at 18:09