<s>[INST] How to prove that $f(x)$ is equal to $ \frac{1}{2}(\exp(x) -\exp(-x)) $? $$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
Since we know that $ \frac{1}{2}(\exp(x) - \exp(-x)) $ is same as 
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{n}}{n!} - \sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}\right)$$
we can place our $f(x)$ in this form. But first we can do more simplifying, we can write that form as $\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n}-(-x)^n}{n!}$. In my book I found next steps, we are placing our $f(x)$ in this last form and we get
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{2n+1}-(-x)^{2n+1}}{(2n+1)!} + \frac{x^{2n}-(-x)^{2n}}{(2n)!}\right)$$
The part which I don't understand is, from where do we get that $\displaystyle\frac{x^{2n}-(-x)^{2n}}{(2n)!}$ ?
 [/INST] They split the sum in an even and an odd part. In general
$$
\sum_{n = 0}^{+\infty} a_n = \sum_{n=0,\;n\; {\rm even}}^{+\infty}a_n + \sum_{n=0,\;n\; {\rm odd}}^{+\infty}a_n = \sum_{n=0}^{+\infty}a_{2n} + \sum_{n=0}^{+\infty}a_{2n+1}
$$
This is particularly useful here because $1-(-1)^{2n} = 1-1 = 0$ and $1 -(-1)^{2n+1} = 1 + 1 = 2$.
So your expression becomes
\begin{eqnarray}
\frac{1}{2}\left(\sum_{n=0}^{+\infty}\frac{x^n}{n!} + \sum_{n=0}^{+\infty}\frac{(-1)^nx^n}{n!}\right) &=& \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^n)x^n}{n!} \\
&=& \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^{2n})x^{2n}}{(2n)!} + \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^{2n+1})x^{2n+1}}{(2n+1)!} \\
&=& 0 + \frac{2}{2}\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!}\\
&=& \sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!}
\end{eqnarray}
</s>