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In order to find the Laurent series of $\frac{1}{\sin{x}}$, I used a well-known formula:

$\sum_{n=0}^{\infty}{x^n} = \frac{1}{1-x} \quad (|x| < 1)$

by

$\frac{1}{\sin{x}} = \frac{1}{z}\frac{1}{1+\left(\sum_{n=1}^{\infty}{\frac{(-1)^n}{(2n+1)!}}z^{2n}\right)} = \frac{1}{z}\sum_{m=0}^{\infty}\left(\sum_{n=1}^{\infty}{\frac{(-1)^n}{(2n+1)!}}z^{2n}\right)^m$

I think this answer is true, but I wonder how I can check the condition of the formula: $|x| < 1$ at this time. Would you be able to lend your expertise?

S tomio
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    The problem is that $|\sin(z)|$ (and, consequently, $|\sin(z)-1|$ ) is unbounded, so the formula does not apply unless you have other domain restrictions. More or less, this is $100$% not the avenue you want to take. – Ninad Munshi Dec 26 '19 at 09:08
  • Your formula converges for $|\sin(x)/x-1|< 1$ ie. for $|x|< r$ where $\sinh(r)/r=1$, when expanding the power and grouping the terms you obtain the correct power series but it converges on the larger domain $|x|< \pi$ (the first zero of $\sin (z)/z$). The Bernouilli polynomials $B_n(t)/n!$ are by definition the coefficients of $\frac{xe^{tx}}{e^x-1}$ – reuns Dec 26 '19 at 11:21

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You could first consider the Taylor expansion of $$\frac{\sin(x)}x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}$$ So $$\frac x{\sin(x)}=1+\frac{x^2}{6}+\frac{7 x^4}{360}+\frac{31 x^6}{15120}+\frac{127 x^8}{604800}+O\left(x^{10}\right)$$ but the problem is to recognize that the coefficients are $$\frac{(-4)^n }{(2 n)!}B_{2 n}\left(\frac{1}{2}\right)$$ where appears Bernoulli polynomials.

Claude Leibovici
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