Well, the title states the problem, so let's get to it.
First of all I've calculated the derivative of the given function:
$$f'(x) = \frac{1}{\sqrt{1+x^2}}$$
Now, we know that:
$$(1+x)^m = 1+\sum_{n=1}^{\infty}{\frac{m\cdot(m-1)\cdot\cdot\cdot(m-n+1)}{n!}\cdot x^n}$$
Therefore:
$$\frac{1}{\sqrt{1+x^2}} = (1+x^2)^{-\frac{1}{2}} = 1+\sum_{n=1}^{\infty}{\frac{(-\frac{1}{2})\cdot(-\frac{3}{2})\cdot \cdot \cdot (-\frac{1}{2}-n+1)}{n!}\cdot x^{2n}} = 1+\sum_{n=1}^{\infty}{\frac{(-1)^n \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \cdot \cdot \frac{2n-1}{2}}{n!}\cdot x^{2n}}$$ Now I'm following a solution from my textbook, and I've managed to follow it up until here. This is where the above expression becomes:
$$=1+\sum_{n=1}^{\infty}{\frac{(-1)^n\cdot(2n-1)!!}{(2n)!!}}\cdot x^{2n}$$
with no explanation given. How was this transformation made? This is in fact the only part of the solution I don't understand, integrating the final sum and thereby getting the power series is no problem, it's just this transformation I can't get through my head.
Any ideas?
Thanks.
Just use the fact that\begin{align}(2n)!!&=(2n)\times(2n-2)\times\cdots\times4\times2\\&=2^n\times n\times(n-1)\times\cdots\times2\times1.\end{align}and that$$(2n-1)!!=(2n-1)\times(2n-3)\times\cdots\times3\times1.$$