# 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.

You may get rid of the fractions by pulling out $$2$$’s from the coefficient of the general term: $$\frac{1/2(1/2-1)...(1/2-n+1)}{n!} =\frac{ (-1)^{n-1}}{2^n n!} (1\times3\times 5\dots (2n-3) )\\ = \frac{(-1)^{n-1}}{2^n n!} \cdot \frac{(2n-2)!}{2\times 4\times \dots \times (2n-2)} \\ =\frac{(-1)^{n-1}(2n-2)!}{2^{2n-1} n!(n-1)!}\\ = \frac{(-1)^{n-1}}{(2n-1)2^{2n-1}} \binom{2n-1}{n}$$
• From $2\times 4\times \dots\times 2(n-1)$, I pulled out $2^{n-1}$. May 22 at 12:57
• What $2$’s remain? I don’t see any other. May 22 at 13:02
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*}