2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Did Feb 1 at 7:16
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\tan\frac{1}{n}=\frac{1}{n}+\frac{1}{3n^3}+\frac{2}{15n^5}+...$ $\endgroup$ – J.Dane Feb 2 at 18:09
  • $\begingroup$ Thank you for the feedback!! $\endgroup$ – Mostafa Ayaz Feb 2 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.