Lets consider the specific example of $$\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}$$here, by the binomial theorem, we've got$$\frac{1}{\sqrt{1+x}}=1+\frac{\left[-\frac{1}{2}\right]}{1!}x+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{2!}x^2+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]\left[-\frac{5}{2}\right]}{3!}x^3+\ldots$$$$=1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\ldots$$ for $|x|<1$. The question is now, is there a way to express this series using summation and Pi notation? The closest I've gotten is the following: $$\frac{1}{\sqrt{1+x}}=1+\sum_{k=0}^a\frac{\prod_{n=0}^a\left[-\frac{1}{2}-n\right]}{(k+1)!}\left(x^{k+1}\right)$$as $a\to\infty$. To some of you the issue might be obvious, but to me it was not as much so. For those of you did not immediately notice, the issue arises with pretty much every value of $a$ apart from $0$, since, if we were to, for example, let $a=2$, then although it is true that we would have our $2$ desired terms, each denominator would be equal to that of the $2^{nd}$ term, which obviously is not what we want. In other words, by the latter equation, we've got, for $a=2$; $$\frac{1}{\sqrt{1+x}}\approx1+\sum_{k=0}^2\frac{\prod_{n=0}^2\left[-\frac{1}{2}-n\right]}{(k+1)!}\left(x^{k+1}\right)$$ $$\approx1+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{1!}x+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{2!}x^2$$ $$\approx1+\frac{3}{4}x+\frac{3}{8}x^2$$which is obviously not true. The question on the table in other words, is whether or not there is a way around this issue?
Any responses are appreciated.
The binomial series can be written in the form $$ (1+x)^{r}=\sum^{\infty}_{k=0}{r\choose k}x^{k} $$ with
$${r\choose k}=\frac{\prod\limits^{k-1}_{j=0}(r-j)}{k!}$$.
With that definition $r$ in the binomial coefficient can be any number, complex, negative, etc
Edit: Extra
$$(1+x)^{-\frac{1}{2}}=\sum^{\infty}_{k=0}{-\frac{1}{2}\choose k}x^{k}=\sum^{\infty}_{k=0}\prod^{k-1}_{s=0}(-\frac{1}{2}-s)\frac{x^{k}}{k!} = $$ $$=\sum^{\infty}_{k=0}\frac{(-1)^{k}}{2^{k}}\prod^{k-1}_{s=0}(1+2s)\frac{x^{k}}{k!}=\sum^{\infty}_{k=0}\frac{(-1)^{k}}{2^{k}}\frac{(2k)!}{2^{k}k!}\frac{x^{k}}{k!}= \sum^{\infty}_{k=0}\frac{(-1)^{k}(2k)!x^{k}}{(k!^{2})4^{k}}= $$ $$=\sum^{\infty}_{k=0}\frac{(-1)^{k}x^{k}}{4^{k}} {2k \choose k} $$