{{begin-eqn}} {{eqn | l = \sin x       | r = 2 \sin \dfrac x 2 \cos \dfrac x 2       | c = Double Angle Formula for Sine }} {{eqn | ll= \leadstoandfrom       | l = \dfrac 1 {\sin x}       | r = \dfrac 1 {2 \sin \dfrac x 2 \cos \dfrac x 2}       | c = taking the reciprocal of both sides }} {{eqn | ll= \leadstoandfrom       | l = \csc x       | r = \dfrac 1 2 \csc \dfrac x 2 \sec \dfrac x 2       | c = Cosecant is Reciprocal of Sine, Secant is Reciprocal of Cosine }} {{eqn | r = \dfrac 1 2 \csc \dfrac x 2 \dfrac 2 {e^{i \frac x 2} + e^{-i \frac x 2} }       | c = Secant Exponential Formulation }} {{eqn | r = \csc \dfrac x 2 \dfrac {e^{-i \frac x 2} } {1 + e^{-i x} } }} {{eqn | ll= \leadstoandfrom       | l = \csc x \paren {1 + e^{-i x} }       | r = \csc \dfrac x 2 e^{-i \frac x 2}       | c = multiplying both sides by $1 + e^{-i x}$ }} {{eqn | ll= \leadstoandfrom       | l = \csc x       | r = \csc \dfrac x 2 e^{-i \frac x 2} - \csc x e^{-i x}       | c = subtracting $\csc x e^{-i x}$ from both sides }} {{eqn | r = \dfrac {2 i} {e^{i \frac x 2} - e^{-i \frac x 2} } e^{-i \frac x 2} - \dfrac {2 i} {e^{i x} - e^{-i x} } e^{-i x}       | c = Cosecant Exponential Formulation }} {{eqn | r = 2 \dfrac i {e^{i x} - 1} - \dfrac {2 i} {e^{2 i x} - 1} }} {{eqn | r = \dfrac 1 x \paren {2 \dfrac {i x} {e^{i x} - 1} - \dfrac {2 i x} {e^{2 i x} - 1} } }} {{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \paren {2 \dfrac {\paren {i x}^n B_n} {n!} - \dfrac {\paren {2 i x}^n B_n} {n!} }       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \dfrac {B_n \paren {2 \paren {i x}^n - \paren {2 i x}^n} } {n!} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {B_n i^n x^{n - 1} 2 \paren {1 - 2^{n - 1} } } {n!} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {B_{2 n} i^{2 n} x^{2 n - 1} 2 \paren {1 - 2^{2 n - 1} } } {\paren {2 n}!}       | c = Odd Bernoulli Numbers Vanish and the term $n = 1$ vanishes }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {B_{2 n} \paren {-1}^{n - 1} x^{2 n - 1} 2 \paren {2^{2 n - 1} - 1} } {\paren {2 n}!} }} {{end-eqn}} {{qed|lemma}} === Convergence === By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \size {\frac {\frac {B_{2 n + 2} \paren {-1}^n x^{2 n + 1} 2 \paren {2^{2 n + 1} - 1} } {\paren {2 n + 2}!} } {\frac {B_{2 n} \paren {-1}^{n - 1} x^{2 n - 1} 2 \paren {2^{2 n - 1} - 1} } {\paren {2 n}!} } }       | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} } {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n - 1} - 1 + 1} {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \size  {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt{\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 2 x^2       | c = Asymptotic Formula for Bernoulli Numbers }} {{eqn | r = \lim_{n \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt{\frac {n + 1} n} \paren {\frac {n + 1} n}^{2 n} } \frac 2 {\pi^2 e^2} x^2 }} {{eqn | r = \lim_{n \to \infty} \size {\paren {\frac {n + 1} n }^{2 n} } \frac 1 {\pi^2 e^2} x^2 }} {{eqn | r = \lim_{n \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 1 {\pi^2 e^2} x^2 }} {{eqn | r = \frac {e^2} {\pi^2 e^2} x^2       | c = {{Defof|Euler's Number|subdef = Limit of Sequence}} }} {{eqn | r = \frac 1 {\pi^2} x^2 }} {{end-eqn}} This is less than $1$ if $\size x < \pi$. Hence by the Ratio Test, the outer radius of convergence is $\pi$ The principal part of the Laurent series is finite so converges for $x \ne 0$. Thus we have the annulus of convergence to be $0 < \size x < \pi$. {{qed}}