# Demonstrate: $\sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}}$

Demonstrate that $\sqrt2$ can be expressed as:

$$\sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}}$$ Where $\left(z\right)_i$ is the Pochhammer symbol $\left(z\right)_i=z(z+1)(z+2)...(z+i-1); (z)_0=1$

This is a nice problem, just wanted to share it.

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Isn't it just a special case of the rule $(1+a)^\alpha=\sum_{i=0}^\infty {\alpha \choose i} a^i$? – tomasz Jan 2 at 4:07
Yes, it is! Going that way may simplifies a lot the process. – Alan Jan 2 at 5:04
I tried the Taylor Series for $\sqrt{x}=\lim_{n\rightarrow \infty}{\sum_{i=0}^{n}\frac{(x-1)^i}{i!}\frac{d^i f(1)}{dx^i}}$, so $\sqrt{2}=\lim_{n\rightarrow\infty}\sum_{i=0}^n \frac{1}{i!}\cdot \frac{d^i f(1)}{dx^i}$. Then, I had to prove that $\frac{d^i f(1)}{dx^i}=\left(-1\right)^i\left(\frac{1}{2}\right)_i$ – Alan Jan 4 at 4:54
@tomasz Could you convert your comment into an answer so that this question gets answered? – user17762 May 15 at 4:25