# Power Series Expansion for Cotangent Function

## Theorem

 $\displaystyle \cot x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^{n} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$ $\displaystyle$ $=$ $\displaystyle \frac 1 x - \frac x 3 - \frac {x^3} {45} - \frac {2 x^5} {945} + \cdots$

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

This converges for $0 < \left|{x}\right| < \pi$.

## Proof

 $\displaystyle \cot x$ $=$ $\displaystyle i \frac {e^{i x} + e^{- i x} } {e^{i x} - e^{- i x} }$ Cotangent Exponential Formulation $\displaystyle$ $=$ $\displaystyle i \frac {e^{2 i x} + 1 } {e^{2 i x} - 1 }$ $\displaystyle$ $=$ $\displaystyle i \left({1 + \frac 2 {e^{2 i x} - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle i + \frac {2 i} {e^{2 i x} - 1}$ $\displaystyle$ $=$ $\displaystyle i + \frac 1 x \frac {2 i x} {e^{2 i x} - 1}$ $\displaystyle$ $=$ $\displaystyle i + \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_n \left({2 i x}\right)^n} {n!}$ Definition of Bernoulli Numbers $\displaystyle$ $=$ $\displaystyle \frac 1 x + \frac 1 x \sum_{n \mathop = 2}^\infty \frac {B_n \left({2 i x}\right)^n} {n!}$ as $B_0 = 1$ and $B_1 = -\dfrac 1 2$ $\displaystyle$ $=$ $\displaystyle \frac 1 x + \frac 1 x \sum_{n \mathop = 1}^\infty \frac {B_{2 n} \left({2 i x}\right)^{2 n} } {\left({2 n}\right)!}$ Odd Bernoulli Numbers Vanish $\displaystyle$ $=$ $\displaystyle \frac 1 x \sum_{n \mathop = 0}^\infty \frac {B_{2 n} \left({2 i x}\right)^{2 n} } {\left({2 n}\right)!}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$

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

 $\displaystyle$  $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\frac {\frac {\left({-1}\right)^n 2^{2 n + 2} B_{2 n + 2} \, x^{2 n + 1} } {\left({2 n + 2}\right)!} } {\frac {\left({-1}\right)^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} } } \right \rvert$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({2 n + 2}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 4 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {B_{2 n + 2} } {B_{2 n} } } \right \rvert 2 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\frac 1 {\left({2 n + 1}\right) \left({n + 1}\right)} \frac {\left({- 1}\right)^{n + 2} 4 \sqrt {\pi (n + 1)} \left({\frac {n + 1} {\pi e} }\right)^{2 n + 2} } {\left({- 1}\right)^{n + 1} 4 \sqrt {\pi n} \left({\frac n {\pi e} }\right)^{2 n} } } \right \rvert 2 x^2$ Asymptotic Formula for Bernoulli Numbers $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\frac {\left({n + 1}\right)^2} {\left({2 n + 1}\right) \left({n + 1}\right)} \sqrt {\frac {n + 1} n } \left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac 2 {\pi^2 e^2} x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\left({\frac {n + 1} n }\right)^{2 n} } \right \rvert \frac {x^2} {\pi^2 e^2}$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left \lvert {\left({\left({1 + \frac 1 n}\right)^n}\right)^2} \right \rvert \frac {x^2} {\pi^2 e^2}$ $\displaystyle$ $=$ $\displaystyle \frac {e^2 x^2} {\pi^2 e^2}$ Definition of Euler's Number $\displaystyle$ $=$ $\displaystyle \frac {x^2} {\pi^2}$

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

$\left|{x}\right| < \pi$

Hence by the Ratio Test, the series converges for $\left|{x}\right| < \pi$.

$\blacksquare$