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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?

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  • $\begingroup$ 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
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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.

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HINT

We have that

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

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

therefore

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

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.

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

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