From Power Series Expansion for Hyperbolic Cotangent Function: :$(1): \quad \coth x = \displaystyle \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$ Then: {{begin-eqn}} {{eqn | l = \tanh x       | r = 2 \coth 2 x - \coth x       | c = Sum of Hyperbolic Tangent and Cotangent }} {{eqn | r = 2 \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, \left({2 x}\right)^{2 n - 1} } {\left({2 n}\right)!} - \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = by $(1)$ }} {{eqn | r =  \sum_{n \mathop = 0}^\infty \frac {2^{2 n} \left({2^{2 n}  - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c =  }} {{eqn | r =  \sum_{n \mathop = 1}^\infty \frac {2^{2 n} \left({2^{2 n}  - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}       | c = the term in $n = 0$ vanishes }} {{end-eqn}} {{qed|lemma}} By Combination Theorem for Limits of Functions we can deduce the following. {{begin-eqn}} {{eqn | o =        | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\frac {2^{2 n + 2} \left({2^{2 n + 2} - 1}\right) B_{2 n + 2} \, x^{2 n + 1} } {\left({2 n + 2}\right)!} } {\frac {2^{2 n} \left({2^{2 n} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} } } \right \rvert       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {\left({2^{2 n + 2} - 1}\right)} {\left({2^{2 n} - 1}\right)} \frac 1 {\left({2 n + 1}\right) \left({2 n + 2}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 4 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\frac {2^{2 n + 2} - 1} {2^{2 n} - 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 2 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 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 2 x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 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 2 x^2       | c = }} {{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 8 x^2       | c = }} {{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 (n + 1)} \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 8 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 8 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac 4 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} \left \lvert {\left({\left({1 + \frac 1 n}\right)^n}\right)^2} \right \rvert \frac 4 {\pi^2 e^2} x^2       | c = }} {{eqn | r = \frac {4 e^2} {\pi^2 e^2} x^2       | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}} }} {{eqn | r = \frac 4 {\pi^2} x^2       | c = }} {{end-eqn}} This is less than $1$ {{iff}}: : $\left|{x}\right| < \dfrac \pi 2$ Hence by the Ratio Test, the series converges for $\left|{x}\right| < \dfrac \pi 2$. {{qed}}