Veamos el ejemplo concreto de $$\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}$$ aquí, por el teorema del binomio, tenemos $$\frac{1}{\sqrt{1+x}}=1+\frac{\left[-\frac{1}{2}\right]}{1!}x+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{2!}x^2+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]\left[-\frac{5}{2}\right]}{3!}x^3+\ldots$$$$ =1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\ldots $$ 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 $; $$ \frac{1}{\sqrt{1+x}}\approx1+\sum_{k=0}^2\frac{\prod_{n=0}^2\left[-\frac{1}{2}-n\right]}{(k+1)!}\left(x^{k+1}\right) $$ $$ \approx1+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{1!}x+\frac{\left[-\frac{1}{2}\right]\left[-\frac{3}{2}\right]}{2!}x^2 $$ $$ \approx1+\frac{3}{4}x+\frac{3}{8}x^2$$ lo que obviamente no es cierto. En otras palabras, la cuestión que está sobre la mesa es si existe o no una forma de evitar este problema.
Agradecemos cualquier respuesta.
