Lists of Taylor and Maclaurin series on this blog

 Here are the lists of Maclaurin series expanded on this blog:

$e^x=\displaystyle\sum_{n=0}^{+\infty} \dfrac{x^{n}}{n!}$  $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\sin(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\sinh(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{x^{2n+1}}{(2n+1)!}$ $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\cos(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}x^{2n}}{(2n)!}$ $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\cosh(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{x^{2n}}{(2n)!}$ $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\dfrac{e^x-1}{x}=\displaystyle\sum_{n=0}^{+\infty} \dfrac{x^n}{(n+1)!}$ $\forall x \in \mathbb{R}$ or $\forall x \in (-\infty,+\infty)$

$\dfrac{x}{e^x-1}=\displaystyle\sum_{n=0}^{+\infty} \dfrac{*B_{n}x^{n}}{n!}$ where $*B_n$ are Bernoulli numbers, $\forall x \in (-2\pi, 2\pi)$

$\dfrac{x}{2}\coth\left(\dfrac{x}{2}\right)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{*B_{2n}x^{n}}{2n!}$ where $*B_{2n}$ are even Bernoulli numbers

$\coth(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}$ $\forall x \in (-\pi, \pi)$

$\cot(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}$ $\forall x \in (-\pi, \pi)$

$\tan(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}(1-2^{n})2^{2n}x^{2n-1}}{(2n)!}=\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^{n-1}B_{2n}(2^{n}-1)2^{2n}x^{2n-1}}{(2n)!}$  $$\forall x \in (-\frac{\pi}{2}, \frac{\pi}{2})$$

$\tanh(x)=\displaystyle\sum_{n=0}^{+\infty} \dfrac{B_{2n}(2^{n}-1)2^{2n}x^{2n-1}}{(2n)!}$ $\forall x \in (-\frac{\pi}{2}, \frac{\pi}{2})$

$\dfrac{1}{1-x}=\displaystyle\sum_{n=0}^{+\infty}x^{n}$ $\forall x \in (-1, 1)$

$\dfrac{1}{1-x^k}=\displaystyle\sum_{n=0}^{+\infty}x^{kn}$ $\forall x \in (-1, 1)$

$\dfrac{1}{1+x}=\displaystyle\sum_{n=0}^{+\infty}(-1)^nx^{n}$ $\forall x \in (-1, 1)$

$\dfrac{1}{1+x^k}=\displaystyle\sum_{n=0}^{+\infty}(-1)^nx^{kn}$ $\forall x \in (-1, 1)$

$\log(1-x)=-\displaystyle\sum_{n=1}^{+\infty} (-1)^{2n+1}\dfrac{x^{n}}{n}$ $\forall x \in (-1, 1)$

$-\log(1-x)=\displaystyle\sum_{n=1}^{+\infty} \dfrac{x^{n}}{n}$ $\forall x \in (-1, 1)$

$\log(1+x)=\displaystyle\sum_{n=1}^{+\infty} (-1)^{n+1}\dfrac{x^{n}}{n}$ $\forall x \in (-1, 1)$

$(a+b)^n=\displaystyle\sum_{i=0}^{+\infty}\binom{n}{i}a^{n-i}bi$

$(1+x)^n=\displaystyle\sum_{i=0}^{+\infty}\binom{n}{i}x^{n} \forall x \in (-1, 1)$