Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite
2

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.

share|improve this question
6  
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

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.