Coefficient of Taylor Series of $\sqrt{1+x}$ The coefficient of $x^3$ in the Taylor series of the function $f(x) = \sqrt{1+x}$ about the point $a = 0$ is $$1\over 16$$ 
Can someone show me how to get this value?

Another way to see this is to expand in Binomial series:
$$
(1+x)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k
$$
Although $\binom{\frac{1}{2}}{k}$ looks intimidating, it's nothing when you replace $\frac{1}{2} = \alpha \Rightarrow \binom{\alpha}{k} = \alpha\cdot (\alpha-1)\cdots(\alpha-k+1) \cdot \frac{1}{k!} = \frac{1}{2} \cdots (-\frac{1}{2}) \cdots \frac{1}{2}-(k-1)\cdot \frac{1}{k!} = \frac{\sqrt{\pi}}{k! 2 \Gamma(\frac{3}{2}-k)}$
This is easy since you only need to plug in $k=3$ and $\Gamma(-\frac{3}{2})=\frac{4 \sqrt{\pi}}{3}$ and therefore $\binom{\frac{1}{2}}{3} = \frac{1}{16}$
