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I am trying to verify for which $z \in \mathbb{R}$ the series $\sum _{n=1}^{\infty } \left(1-\cos \left(\frac{1}{n}\right)\right)^z$ converges. The only test that was successful for me is the Kummer Test which gave the apparently correct result that it converges if $z > \frac{1}{2}$.

To get there I use the fact that the series $\sum _{n=1}^{\infty } a_n$ converges (when the limit exists) if $\lim_{n\to \infty } \, n \left(\frac{a_n}{a_{n+1}}-1\right) > 1$

Using Mathematica I get $\lim_{n\to \infty } \, n \left(\frac{a_n}{a_{n+1}}-1\right)=\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos \left(\frac{1}{n+1}\right)\right)^{-z}-1\right) = 2z$ and therefore $z > \frac{1}{2}$.

Start reading here if you are only interested in the problem and not how i got there:

Now I try to understand how to get that

$\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos \left(\frac{1}{n+1}\right)\right)^{-z}-1\right)=\lim_{x\to 0} \, \frac{-1+(1-\cos (x))^z \left(1-\cos \left(\frac{x}{1+x}\right)\right)^{-z}}{x} = 2z$

All attempts I did to show this in an elementary way failed and thats why I hope someone has an idea how to verify the result. For me its not obvious why the result to the limit is how it is

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For anyone reading this question, the assertion that $\sum _{n=1}^{\infty } a_n$ converges if and only $\lim_{n\to \infty } \, n \left(\frac{a_n}{a_{n+1}}-1\right) > 1$ is not true. The inequality forces the series to converge, but the converse is not true, as taking $a_n=1/n^2$ shows: it gives a convergent series where the limit in the inequality equals to zero. – Martin Argerami Feb 21 '12 at 21:46
If you want, I can delete the comment, then. – Martin Argerami Feb 21 '12 at 22:04
up vote 2 down vote accepted

The key is to use the Taylor series approximation $$\cos x \approx 1 - \frac{x^2}{2},$$ which is good for small $x$. That applies to both $x$ and $x/(1+x)$, and so $$\frac{1-\cos x}{1-\cos \frac{x}{1+x}} \approx (1+x)^2.$$ Substituting this into your expression, we get that the expression is $$\approx \frac{(1+x)^{2z}-1}{x} \approx \frac{(1+2zx)-1}{x} = 2z.$$ This reasoning can be made formal using big-O (or little-O) notation, but I leave that for you to ponder.

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Note that ${\displaystyle \lim_{x \rightarrow 0} {1 - \cos(x) \over x^2} = {1 \over 2}}$, so that ${\displaystyle \lim_{n \rightarrow \infty} {\big(1 - \cos({1 \over n})\big)^z \over ({1 \over n^2})^z} = {1 \over 2^z}}$

Thus you can use the limit comparison test with ${\displaystyle\sum_{n=1}^{\infty} \bigg({1 \over n^2}\bigg)^z = \sum_{n=1}^{\infty} {1 \over n^{2z}}}$. This converges iff $z > {1 \over 2}$ by the $p$-test.

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Ok thanks :-). So my original problem is solved. I would still be interested how the limit could be solved and if my approach would also work, even if your way is a lot more elegant and simple. – Listing Jan 1 '11 at 22:22

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