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Determine the set of real numbers x for which the following series diverges

$$\sum_{n=1}^{\infty}\left(\frac{1}{n}\csc\frac{1}{n}-1\right)^x$$

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Sorry for the ignorance, but what does CSC means? –  ofer Nov 30 '11 at 8:40
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@ofer: cosecant, $$\csc (x) = \frac{1}{\sin (x)}.$$ –  InterestedGuest Nov 30 '11 at 8:42

2 Answers 2

One has

$${\sin z\over z}=1-{z^2\over 6}g(z)\ ,$$

where $g$ is analytic in a neighborhood of $0$ and $g(0)=1$. It follows that

$${z\over\sin z} -1 ={1 \over 1 -z^2 g(z)/6}-1 ={z^2 g(z)/6 \over 1 -z^2 g(z)/6} = {z^2\over 6} h(z)\ ,$$

where $h$ is analytic in a neighborhood of $0$ and $h(0)=1$. Therefore your sum can be written as $\sum_{n=1}^\infty a_n$ with

$$a_n\ :=\ \left({1\over 6 n^2} h\bigl({1\over n}\bigr)\right)^x\ .$$

As $h(0)=1$ there is an $n_0$ such that ${1\over2} \leq h\bigl({1\over n}\bigr)\leq 2$ $\ (n>n_0)$ and therefore

$${1\over2^{|x|}} \leq \Bigl(h\bigl({1\over n}\bigr)\Bigr)^x\leq 2^{|x|}\qquad(n>n_0)\ .$$

It follows that

$${1\over 6^x\ 2^{|x|}}\ {1\over n^{2x}}\ \leq\ a_n\ \leq \ {2^{|x|}\over6^x} \ {1\over n^{2x}}\qquad(n>n_0)\ .$$

From this we conclude that the given series is divergent when $x\leq{1\over2}$, and convergent when $x>{1\over2}$.

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Cauchy n-th root test would be the easiest tool here.

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