x=(1/(1+(1/(2+(1/(3+(1/(4+...))))))))=?


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Question Number 45356 by MrW3 last updated on 12/Oct/18

$x = \frac{1}{1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4 + . . .}}}} = ?$
	Answered by MJS last updated on 12/Oct/18
	$non−periodic infinite continued fractions are transcendental numbers, at least I seem to remember this. we can only approximate in this case the sequence is 1, (2/3), (7/(10)), ((30)/(43)), ((157)/(224)), ((972)/(1393)), ((6961)/(9976)), ((56660)/(81201)), ((516901)/(740785)), ((5225670)/(7489051)), ... the difference between the last two fractions is (1/(5547776645035))<10^(−12) I get x≈.697774657964$
	$non - periodic infinite continued fractions$ $are transcendental numbers, at least I seem$ $to remember this . we can only approximate$ $in this case the sequence is$ $1, \frac{2}{3}, \frac{7}{10}, \frac{30}{43}, \frac{157}{224}, \frac{972}{1393}, \frac{6961}{9976}, \frac{56660}{81201}, \frac{516901}{740785}, \frac{5225670}{7489051}, . . .$ $the difference between the last two fractions$ $is \frac{1}{5547776645035} < 10^{- 12}$ $I get x \approx . 697774657964$
	Commented by MrW3 last updated on 12/Oct/18

	$T h a n k y o u f o r t h i s a n s w e r s i r!$
	Commented by MJS last updated on 12/Oct/18

	$Sir, may I offer you these as a gift :$ $these are by your special friend Lambert :$ $\frac{4}{π} = 1 + \frac{1^{2}}{3 + \frac{2^{2}}{5 + \frac{3^{2}}{7 + \frac{4^{2}}{9 + \frac{5^{2}}{⋱}}}}}$ $\tan x = \frac{x}{1 - \frac{x^{2}}{3 - \frac{x^{2}}{5 - \frac{x^{2}}{7 - \frac{x^{2}}{⋱}}}}}$ $these are by Euler :$ $e - 2 = \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{4 + \frac{1}{1 + \frac{1}{1 + \frac{1}{6 + \frac{1}{⋱}}}}}}}}} = \frac{1}{1 + \frac{1}{2 + \frac{2}{3 + \frac{3}{4 + \frac{4}{5 + \frac{5}{6 + \frac{6}{7 + \frac{7}{8 + \frac{8}{⋱}}}}}}}}}$
	Commented by MrW3 last updated on 13/Oct/18

	$v e r y i n t e r e s t i n g, t h a n k y o u s i r .$
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