Taylor series of $\sqrt{1+x}$ I have to give the Taylor series of $\sqrt{1+x}$ with development point $0$.
But I am not quite sure what to do.
Normally you have to give the Taylor series up to a certain order.
Well $T_f(x,0)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$
I calculated the derivative of $f(x)=\sqrt{1+x}$ up to order 5 and locked for a regularity. 
$f^{(n)}(0)=(-1)^n\cdot\frac{\prod_{k\leq n\,\text{with}\, k\,\text{odd}}k}{2^{n}}$
Then the Taylor series is given by:
$T_{f}(x,0)=\sum_{n=0}^\infty \frac{(-1)^n}{\prod_{1\leq k\leq n\,\text{with}\, k\,\text{even}}k\cdot 2^n}x^n$
Which is pretty ugly to work with.
Can you confirm this solution?
Thanks in advance.

Almost correct:
$$\begin{align*}
f^{(0)}(x) & = (1+x)^{\frac{1}{2}} \\[1ex]
f^{(1)}(x) & = \frac{1}{2} \cdot (1+x)^{-\frac{1}{2}} \\[1ex]
f^{(2)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot (1+x)^{-\frac{3}{2}} \\[1ex]
\vdots \\
f^{(n)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot \ldots \cdot \left( - \frac{2n-3}{2} \right) \cdot (1+x)^{-\frac{2n-1}{2}} \\[1ex]
f^{(n)}(0) & = \frac{(-1)^{\color{red}{n-1}}}{2^n} \cdot \prod_{\substack{k \leqslant \color{red}{2n-3} \\ k \text{ odd}}} k \cdot (1+x)^{-\frac{2n-1}{2}}
\end{align*}$$
Usually a convenient notation is used: for $\mu \in \mathbb{R}$ and $n \in \mathbb{N}$ let 
$$\binom{\mu}{n} = \frac{\mu \cdot (\mu-1) \cdot \ldots \cdot (\mu - n + 1)}{n!}.$$
Then the Taylor series expresses nicely as 
$$T_f(x, 0) = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n.$$
