Originally Posted by
simplependulum wow !
How can people ( is that you ?) find out this this expression/series for irrational numbers ?
we have
$\displaystyle x_1 = 3 , x_2 = 7 , x_3 = 47 , x_4 = 2207 , x_5 = 4870847 ... $
we can see that the series converges so quickly !
for this problem ,
I use a quite complicated method , is there any method better than mine ?
Go through it !
the general term of the seqence is $\displaystyle 2 \cosh(2^{n-1}y) $
where $\displaystyle y = \cosh^{-1}(x_1 /2 ) $
so the product of first $\displaystyle n $ terms is
$\displaystyle x_1 x_2 x_3 .. x_n = $
$\displaystyle 2^n \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $
$\displaystyle = \frac{2^n}{\sinh(y)} \cdot \sinh(y) \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y) $
$\displaystyle = \frac{2^{n-1}}{\sinh(y)} \cdot \sinh(2y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$
$\displaystyle = \frac{2^{n-2}}{\sinh(y)} \cdot \sinh(4y) \cosh(4y) ... \cosh(2^{n-1}y) = \frac{\sinh(2^n y)}{ \sinh(y) } $
the series can be written in this form :
$\displaystyle 2\sinh(y) \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$
$\displaystyle \sqrt{x_1^2 - 4 } \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$
To evluate this series , we just need to create $\displaystyle \frac{1}{a^2 - 1} - \frac{1}{a^2 - 1} $ next to the series .
I got
$\displaystyle \sqrt{ x_1^2 - 4 } \left( \frac{1}{ e^{2y} - 1} - \frac{1}{ e^{ 2^{m+1} y } - 1 } \right ) $
$\displaystyle e^y = \frac{ x_1 + \sqrt{ x_1^2 - 4 }}{2} $
for $\displaystyle x_1 = 3 , m \to \infty $
I got $\displaystyle \frac{ 3 - \sqrt{5} }{2} $
or an elegant-appearance
$\displaystyle \sqrt{5} = 3 - \frac{2}{3} - \frac{2}{3\cdot7} - \frac{2}{3 \cdot 7 \cdot 47 } - \frac{2}{3\cdot7\cdot47\cdot2207} - .. $
the fifth term is about $\displaystyle 10^{-6} $ , wow