Deriving the power series for $\tan(x)$ and $\tanh(x)$

Some times ago, we have been able to obtain the power series of $\tan(x)$ and $\tanh(x)$. But this method of long division fails to make visible the connection between the power series of $\tan(x)$ and $\tanh(x)$ and the Bernoulli numbers. It also comes quite unhandy. In the previous post, we have successfully derived the formula for $\cot(x)$ and $\coth(x)$. We are now ready to derive the general formulae of the power series of $\tan(x)$ and $\tanh(x)$

The crucial identities that we are going to use to derive the series are:

$$\tan(x)=\cot(x)-2\cot(2x)\tag{1}$$

$$\tanh(x)=2\coth(2x)-\coth(x)\tag{2}$$

The power series that we need to use are

$$\cot(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{3}$$

$$\coth(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4}$$

$$2\cot(2x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n+1}2^{2n-1}x^{2n-1}}{(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{4n}x^{2n-1}}{(2n)!}\tag{5}$$

$$2\coth(2x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n+1}2^{2n-1}x^{2n-1}}{(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{4n}x^{2n-1}}{(2n)!}\tag{6}$$

Power series of $\tan(x)$

We apply the identity from $(1)$

$$\tan(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}-\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{4n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}-[(-1)^{n}B_{2n}2^{4n}x^{2n-1}]}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}[2^{2n}x^{2n-1}-2^{4n}x^{2n-1}]}{(2n)!}$$

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}(1-2^{2n})}{(2n)!}}$$

or

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n-1}B_{2n}2^{2n}x^{2n-1}(2^{2n}-1)}{(2n)!}}$$

Power series of $\tanh(x)$

$$\tanh(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{4n}x^{2n-1}}{(2n)!}-\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{B_{2n}2^{4n}x^{2n-1}-B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{B_{2n}[2^{4n}x^{2n-1}-2^{2n}x^{2n-1}]}{(2n)!}$$

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{B_{2n}2^{2n}x^{2n-1}(2^{2n}-1)}{(2n)!}}$$