Power Series Expansion for Hyperbolic Cotangent Function

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Theorem

The hyperbolic cotangent function has a Taylor series expansion:

\(\ds \coth x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \frac 1 x + \frac x 3 - \frac {x^3} {45} + \frac {2 x^5} {945} - \frac {x^7} {4725} + \cdots\)


where $B_{2 n}$ denotes the Bernoulli numbers.

This converges for $0 < \size x < \pi$.


Proof

\(\ds \coth x\) \(=\) \(\ds \frac {e^x + e^{-x} } {e^x - e^{-x} }\) Definition 1 of Hyperbolic Cotangent
\(\ds \) \(=\) \(\ds \frac {e^{2 x} + 1} {e^{2 x} - 1}\)
\(\ds \) \(=\) \(\ds 1 + \frac 2 {e^{2 x} - 1}\)
\(\ds \) \(=\) \(\ds 1 + \frac 1 x \frac {2 x} {e^{2 x} - 1}\)
\(\ds \) \(=\) \(\ds 1 + \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_n \paren {2 x}^n} {n!}\) Definition of Bernoulli Numbers
\(\ds \) \(=\) \(\ds \frac 1 x + \frac 1 x \sum_{n \mathop = 2}^\infty \frac {B_n \paren {2 x}^n} {n!}\) as $B_0 = 1$ and $B_1 = -\dfrac 1 2$
\(\ds \) \(=\) \(\ds \frac 1 x + \frac 1 x \sum_{n \mathop = 1}^\infty \frac {B_{2 n} \paren {2 x}^{2 n} } {\paren {2 n}!}\) Odd Bernoulli Numbers Vanish
\(\ds \) \(=\) \(\ds \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_{2 n} \paren {2 x}^{2 n} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}\)


By Combination Theorem for Limits of Real Functions we can deduce the following:

\(\ds \) \(\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {\frac {2^{2 n + 2} B_{2 n + 2} \, x^{2 n + 1} } {\paren {2 n + 2}!} } {\frac {2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} } }\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi (n + 1)} \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 2 x^2\) Asymptotic Formula for Bernoulli Numbers
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 2 {\pi^2 e^2} x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac {x^2} {\pi^2 e^2}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac {x^2} {\pi^2 e^2}\)
\(\ds \) \(=\) \(\ds \frac {e^2 x^2} {\pi^2 e^2}\) Definition of Euler's Number as Limit of Sequence
\(\ds \) \(=\) \(\ds \frac {x^2} {\pi^2}\)

This is less than $1$ if and only if:

$\size x < \pi$

Hence by the Ratio Test, the series converges for $\size x < \pi$.

$\blacksquare$


Also see


Sources

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