{{begin-eqn}} {{eqn | l = \sinh x       | r = 2 \sinh \dfrac x 2 \cosh \dfrac x 2       | c = Double Angle Formula for Hyperbolic Sine }} {{eqn | ll= \iff       | l = \dfrac 1 {\sinh x}       | r = \dfrac 1 {2 \sinh \dfrac x 2 \cosh \dfrac x 2}       | c = taking the reciprocal of both sides }} {{eqn | ll= \iff       | l = \csch x       | r = \dfrac 1 2 \csch \dfrac x 2 \sech \dfrac x 2       | c = Cosecant is Reciprocal of Sine, Secant is Reciprocal of Cosine }} {{eqn | r = \dfrac 1 2 \csch \dfrac x 2 \dfrac 2 {e^{\frac x 2} + e^{\frac x 2} }       | c = {{Defof|Hyperbolic Secant|index = 1}} }} {{eqn | r = \csch \dfrac x 2 \dfrac {e^{-\frac x 2} } {1 + e^{-x} } }} {{eqn | ll= \iff       | l = \csch x \left({1 + e^{-x} }\right)       | r = \csch \dfrac x 2 e^{-\frac x 2}       | c = multiplying both sides by $1 + e^{-x}$ }} {{eqn | ll= \iff       | l = \csch x       | r = \csch \dfrac x 2 e^{-\frac x 2} - \csch x e^{-x}       | c = subtracting $\csch x e^{-x}$ from both sides }} {{eqn | r = \dfrac 2 {e^{\frac x 2} - e^{-\frac x 2} } e^{-\frac x 2} - \dfrac 2 {e^x - e^{-x} } e^{-x}       | c = {{Defof|Hyperbolic Cosecant|index = 1}} }} {{eqn | r = 2 \dfrac 1 {e^x - 1} - \dfrac 2 {e^{2 x} - 1} }} {{eqn | r = \dfrac 1 x \left({2 \dfrac x {e^x - 1} - \dfrac {2 x} {e^{2 x} - 1} }\right) }} {{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \left({2 \dfrac {x^n B_n} {n!} - \dfrac {\left({2 x}\right)^n B_n} {n!} }\right)       | c = {{Defof|Bernoulli Numbers}} }} {{eqn | r = \dfrac 1 x \sum_{n \mathop = 0}^\infty \dfrac {B_n \left({2 x^n - \left({2 x}\right)^n}\right)} {n!} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {B_n x^{n - 1} 2 \left({1 - 2^{n - 1} }\right)} {n!} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {2 \left({1 - 2^{2 n - 1} }\right) B_{2 n} x^{2 n - 1} } {\left({2 n}\right)!}       | c = Odd Bernoulli Numbers Vanish and the term $n = 1$ vanishes }} {{end-eqn}} {{qed}} === Convergence === By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \left\lvert {\frac {\frac {B_{2 n + 2} x^{2 n + 1} 2 \left({2^{2 n + 1} - 1}\right)} {\left({2 n + 2}\right)!} } {\frac {B_{2 n} x^{2 n - 1} 2 \left({2^{2 n - 1} - 1}\right)} {\left({2 n}\right)!} } }\right\rvert       | r = \lim_{n \mathop \to \infty} \left\lvert {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} \frac 1 {\left({2 n + 1}\right) \left({2 n + 2}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} }\right\rvert \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac {2^{2 n + 1} } {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} }\right\rvert \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {4 \frac {2^{2 n - 1} - 1 + 1} {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} }\right\rvert \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {4 + \frac 4 {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} }\right\rvert \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert \frac 1 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } }\right\rvert 2 x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {\left({-1}\right)^{n + 2} 4 \sqrt{\pi \left({n + 1}\right)} \left({\frac {n + 1} {\pi e} }\right)^{2 n + 2} } {\left({-1}\right)^{n + 1} 4 \sqrt {\pi n} \left({\frac n {\pi e} }\right)^{2 n} } }\right\rvert 2 x^2       | c = Asymptotic Formula for Bernoulli Numbers }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\frac {\left({n + 1}\right)^2} {\left({2 n + 1}\right) \left({n + 1}\right)} \sqrt{\frac {n + 1} n} \left({\frac {n + 1} n}\right)^{2 n} }\right\rvert \frac 2 {\pi^2 e^2} x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\left({\frac {n + 1} n }\right)^{2 n} }\right\rvert \frac 1 {\pi^2 e^2} x^2 }} {{eqn | r = \lim_{n \mathop \to \infty} \left\lvert {\left({\left({1 + \frac 1 n}\right)^n}\right)^2}\right\rvert \frac 1 {\pi^2 e^2} x^2 }} {{eqn | r = \frac {e^2} {\pi^2 e^2} x^2       | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}} }} {{eqn | r = \frac 1 {\pi^2} x^2 }} {{end-eqn}} This is less than $1$ if $\left\lvert{x}\right\rvert < \pi$. 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 $0 < \left\lvert{x}\right\rvert < \pi$. {{qed}}