taylor expansion of $\sinh(x)$ 
I would like to find taylor expansion of $sh(x)$ 

My thoughts
indeed, 
note that : $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ then
\begin{align}
\sinh(x)&=\frac{e^x-e^{-x}}{2} \\
&=\frac{1}{2}\left( e^x-e^{-x} \right)\\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^n\frac{x^k}{k!}+o(x^n)-\sum_{k=0}^n\frac{(-1)^k x^{k}}{k!}-o(x^n)\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{x^{k}}{k!}-\frac{(-1)^{k}x^{k}}{k!}\right)+o(x^{n})\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right)\\
&\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{n}\left( \dfrac{(1+(1)^{2(k'+1)})x^{2k'+1}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{n}\left( \dfrac{(1+(-1)^{2k'+1})x^{k}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\
&\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{2x^{2k'+1}}{k!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{2n'}\left( \dfrac{(1-1)x^{2k'}}{2k'!}\right)+o(x^{2n'})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\ &\underset{x\to 0}=\begin{cases}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{x^{2k'+1}}{(2k'+1)!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\ 0 & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}
\end{align} 
Update


*

*but if I want from there


$$\sinh(x)\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + 
  \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n}) $$


*

*to get the desired result:


$$\sinh(x)\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right)$$
indeed,
\begin{align}
 sh(x)&\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + 
   \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
 &\underset{x\to 0}=0 + 
   \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{2)x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
 &\underset{x\to 0}=
   \sum_{k'=0}^{E((n-1)/2)} \frac{x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
 &\underset{x\to 0}=\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\\
 &\text{These few passing I would like to know them. }\\
 sh(x)&\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right)
 \end{align}


*

*Is my proof correct 

*I'm interested in more ways of finding taylor expansion of $\sinh(x)$.

 A: How about a rather simple derivation like the one below: 
$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$
and 
$$\exp(-x)= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
So when you subtract the two equations: 
$$\exp(x) - \exp(-x) = 2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!}$$
Finally, 
$$\frac{\exp(x) - \exp(-x)}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{(2n+1)!}$$
Odd powers remain and sine is an odd function. 
A: There are a couple things wrong here.
The big sum has not two values according to the even-ness of $k$, as it does not depend on any variable called $k$. And in those sums, the upper bound value is false.

You really wanted to write:
\begin{align}
S
&=\frac{1}{2}\sum_{k=0}^{n} \frac{(1+(-1)^{k+1})x^{k}}{k!}+o(x^{n}) \\
&=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + 
  \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} +
o(x^{n}) \\
\end{align}
