Wallis' product exploration

02092020, 01:57 PM
Post: #10




RE: Wallis' product exploration
Using identity 5.36 ( p. 186 of Concrete Mathematics )
\( \binom{n\frac{1}{2}} {n} = \binom{2n} {n} ÷ 2^{2n} \) I think we can also reduce the limit to: \( \pi = \lim\limits_{n\to\infty} \frac {1} {n+\frac {1}{2}} ÷ \binom{n\frac{1}{2}} {n}^{2} \) (I could have messed something up there..) 17bii  32s  32sii  41c  41cv  41cx  42s  48g  48g+  48gx  50g  30b 

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