The convergence of the following series as $a \in \mathbb{R}$

$$\sum_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{k(n)}\,\,\,\,; k(n)=\frac{1}{\sin^a\left(\frac{1}{n}\right)}$$

As $n \to +\infty$ we have that $\cos\frac{1}{n} \sim 1-\frac{1}{2n^2}$ and that $\sin^a\left(\frac{1}{n}\right) \sim \frac{1}{n^a}$ but I can't figure out how to handle these results. Better to use comparision test?

  • $\begingroup$ Yes. You could use those familiar Maclaurin formula to estimate the general term. For the $x^y$ type, you may find it useful to consider $\exp(x \log(y))$ instead. $\endgroup$ – xbh Aug 16 '18 at 7:46
  • $\begingroup$ Tip on MathJax: \; would give a thick space. If possible \quad would give a thicker space. $\endgroup$ – xbh Aug 16 '18 at 7:56

First you can have: $\left(\cos\frac{1}{n}\right)^{\frac{1}{\sin^a\left(\frac{1}{n}\right)}}=e^{\frac{1}{\sin^a\left(\frac{1}{n}\right)}\log\cos\frac{1}{n}}.$

And $\frac{1}{\sin^a\left(\frac{1}{n}\right)}\log\cos\frac{1}{n}=\frac{1}{\sin^a\left(\frac{1}{n}\right)}\log(1+\cos\frac{1}{n}-1)\sim\frac{\cos\frac{1}{n}-1}{\frac{1}{n^a}}\sim\frac{\frac{-1}{2n^2}}{\frac{1}{n^a}}=\frac{-1}{2n^{2-a}},$ as $n\to\infty.$ So $$\left(\cos\frac{1}{n}\right)^{\frac{1}{\sin^a\left(\frac{1}{n}\right)}}\sim e^{\frac{-1}{2}n^{a-2}}.$$ Base on this, you can give the convergence.



We have that

$$\cos\frac{1}{n} \sim 1-\frac{1}{2n^2}$$

$$\sin^a\left(\frac{1}{n}\right) \sim \frac{1}{n^a}$$


$$\left(\cos\frac{1}{n}\right)^{k(n)}=e^{k(n)\log \left(\cos\frac{1}{n}\right)}\sim e^{-\frac12 n^{a-2}}$$

and the given series seems to converge for $a>2$.

Now you can try to make it more rigorous taking into account the remainders and referring to limit comparison test.

  • $\begingroup$ It was a copy and paste of the OP! $\endgroup$ – gimusi Aug 16 '18 at 7:55

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.