# Power Series Expansion for Tangent Function/Proof 2

## Theorem

 $\displaystyle \tan x$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n - 1} 2^{2 n} \left({2^{2 n} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$ $\displaystyle$ $=$ $\displaystyle x + \frac {x^3} 3 + \frac {2 x^5} {15} + \frac {17 x^7} {315} + \cdots$

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

This converges for $\left|{x}\right| < \dfrac \pi 2$.

## Proof

We have:

 $\displaystyle \frac x {e^x - 1}$ $=$ $\displaystyle \frac x 2 \left({\frac 2 {e^x - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac x 2 \left({\frac {e^x - e^x + 2} {e^x - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)$ $\displaystyle$ $=$ $\displaystyle -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle -\frac x 2 + \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)$ multiplying top and bottom by $e^{-x/2}$

Thus:

 $(1):\quad$ $\displaystyle \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} x^{2 n}$ Definition of Bernoulli numbers

Replacing $x$ with $2 i x$ in the left hand side $(1)$:

 $\displaystyle$  $\displaystyle \frac {2 i x} 2 \left({\frac {e^{2 i x / 2} + e^{-2 i x / 2} } {e^{2 i x / 2} - e^{-2 i x / 2} } }\right)$ $\displaystyle$ $=$ $\displaystyle i x \left({\frac {e^{i x} + e^{-i x} } {e^{i x} - e^{-i x} } }\right)$ $\displaystyle$ $=$ $\displaystyle x \cot x$ Cotangent Exponential Formulation

Replacing $x$ with $2 i x$ in the right hand side $(1)$:

 $\displaystyle$  $\displaystyle \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} \left({2 i x}\right)^{2 n}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}$ $\displaystyle \leadsto \ \$ $\displaystyle x \cot x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}$ $\displaystyle \leadsto \ \$ $\displaystyle \cot x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n - 1}$

Then from Cotangent Minus Tangent:

$\tan x = \cot x - 2 \cot 2 x$

from which:

 $\displaystyle \tan x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1} - 2 \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} \left({2 x}\right)^{2 n - 1}$ by $(1)$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} (1 - 2^{2 n}) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^{n - 1} 2^{2 n} (2^{2 n} - 1) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}$

$\Box$

### Proof of Convergence

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

 $\displaystyle$  $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {\frac {\paren {-1}^n 2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} } {\paren {2 n + 2}!} x^{2 n + 1} } {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1} } }$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 2} - 1} {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } { B_{2 n} } } 8 x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 8 x^2$ Asymptotic Formula for Bernoulli Numbers $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 8 {\pi^2 e^2} x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 4 {\pi^2 e^2} x^2$ $\displaystyle$ $=$ $\displaystyle \frac {4 e^2} {\pi^2 e^2} x^2$ Definition of Euler's Number $\displaystyle$ $=$ $\displaystyle \frac 4 {\pi^2} x^2$

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

$\size x < \dfrac \pi 2$

Hence by the Ratio Test, the series converges for $\size x < \dfrac \pi 2$.

$\blacksquare$