Theorem: Power Series Expansion for Hyperbolic Tangent Function

The hyperbolic tangent function has a Taylor series expansion:
{{begin-eqn}}
{{eqn | l = \tanh x
      | r = \sum_{n \mathop = 1}^\infty \frac {2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = x - \frac {x^3} 3 + \frac {2 x^5} {15} - \frac {17 x^7} {315} + \frac {62 x^9} {2835} - \cdots
      | c = 
}}
{{end-eqn}}
where $B_{2 n}$ denotes the Bernoulli numbers.
This converges for $\size x < \dfrac \pi 2$.

Proof:
From Power Series Expansion for Hyperbolic Cotangent Function:
:$(1): \quad \coth x = \ds \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}$
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} \, \paren {2 x}^{2 n - 1} } {\paren {2 n}!} - \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = by $(1)$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \frac {2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = the term in $n = 0$ vanishes
}}
{{end-eqn}}
{{qed|lemma}}
By Combination Theorem for Limits of Real Functions we can deduce the following.
{{begin-eqn}}
{{eqn | o = 
      | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} \, x^{2 n + 1} } {\paren {2 n + 2} !} } {\frac {2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} } }
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 2} - 1} {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
      | c =
}}
{{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} } } 8 x^2
      | c =
}}
{{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 (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} } } 8 x^2
      | c = Asymptotic Formula for Bernoulli Numbers
}}
{{eqn | r = \lim_{n \mathop \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 8 {\pi^2 e^2} x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2
      | c =
}}
{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \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 as Limit of Sequence}}
}}
{{eqn | r = \frac 4 {\pi^2} x^2
      | c =
}}
{{end-eqn}}
This is less than $1$ {{iff}}:
:$\size x < \dfrac \pi 2$
Hence by the Ratio Test, the series converges for $\size x < \dfrac \pi 2$.
{{qed}}