Stoneham's series

[Albert Chan]

(06-09-2024 10:36 PM)Gerson W. Barbosa Wrote:  \[\pi \approx \left(4\prod _{k=1}^{n}{\frac {(2k+1)^2-1}{(2k+1)^2}}\right) \left({1-\frac{2}{8n+9+\frac{1\times3}{8n+8+\frac{3\times5}{8n+8+… +\frac{\left({2n}\right)^2-1}{8n+8}}}}}\right)\]

Nice!

I have trouble doing correction for √2 ... it just seems too complicated.

\(\sqrt{2} \approx \displaystyle
\left(2\prod _{k=1}^{n} \left( 1 - \frac {1/4}{(2k-1)^2} \right) \right)
\left({1-\frac{2}
{1+32n+\frac{(4^2-1)^2/(2^2-1)}
{32n+\frac{(8^2-1)^2/(4^2-1)}
{32n+… +\frac{{((4n)^2-1)^2/((2n)^2-1)}}{32n}}}}}
\right)\)