Binomial coefficient and series expansion I had to expand $y=\sqrt{1+x^2}$ into the Maclaurin series, which (prior to simplification) would give me: \begin{align} \sqrt{1+x^2}= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}+...+\frac{1/2(1/2-1)...(1/2-n+1)x^{2n}}{n!}+...\end{align}
What is the nice way to compactify the series using the $\sum$ notation? I need that because I will have to check the convergence of the series at $x=-1$ and $x=1$ (at which the series seems to converge). 

From my lecture notes I can see that $(1+x)^m=\sum^{\infty}_{n=0} C^{n}_{m}x^n,|x|<1$, but because we've never used the combination notion, I'm confused as to what the correct way to implement it is. 

It appears that I can multiply the numerator and the denominator of my fraction $\frac{1/2(1/2-1)...(1/2-n+1)x^{2n}}{n!}$ by $(1/2-n)!$ to turn it into $\frac {1/2!}{n!(1/2-n)!}$, which seems to be the normal expansion for the notion $C^n_m, m=1/2$.


However, would it be right of me to use the factorial of a fraction here or am I mixing some incompatible notations? Is there something I'm doing completely wrong? I would be grateful for a clarification.
 A: You may get rid of the fractions by pulling out $2$’s from the coefficient of the general term: $$ \frac{1/2(1/2-1)...(1/2-n+1)}{n!} =\frac{ (-1)^{n-1}}{2^n n!} (1\times3\times 5\dots (2n-3) )\\ = \frac{(-1)^{n-1}}{2^n n!} \cdot \frac{(2n-2)!}{2\times 4\times \dots \times (2n-2)} \\ =\frac{(-1)^{n-1}(2n-2)!}{2^{2n-1} n!(n-1)!}\\ = \frac{(-1)^{n-1}}{(2n-1)2^{2n-1}} \binom{2n-1}{n}$$
A: A convenient generalisation of the binomial coefficient is given for real (or even complex) values $\alpha$ by
\begin{align*}
\binom{\alpha}{n}:=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}&n\geq 0\tag{1}\\
0&n<0
\end{cases}
\end{align*}
Using (1) we can write
\begin{align*}
\sqrt{1+x^2}&= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}\\
&\qquad +\cdots+\frac{1/2(1/2-1)\cdots(1/2-n+1)x^{2n}}{n!}+\cdots\\
&= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}\\
&\qquad +\cdots+\color{blue}{\binom{\frac{1}{2}}{n}}x^{2n}+\cdots\\
&=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n}x^{2n}
\end{align*}
The formula (1) can be found for instance as formula (5.1) in Concrete Mathematics by R. L. Graham, D. Knuth and O. Patashnik.
