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Question Number 124870 by Algoritm last updated on 06/Dec/20

	Answered by mindispower last updated on 06/Dec/20

	$l n (x_{n + 1}) = \frac{1}{4} l n (81) - \frac{1}{4} l n (x_{n})$ $W_{n} = l n (x_{n})$ $W_{n + 1} = l n (3) - \frac{w_{n}}{4}$ $f i n d w_{n} w i t h e n t h a n x_{n} = e^{w_{n}}$ $s o l v e x_{n} ⩽ 3 t h e a n s w e r w i l l b e e n \in [α, + \infty [$ $n_{0} = [α] + 1$
	Commented by Algoritm last updated on 06/Dec/20

	$W_{n} - lambert ?$
	Answered by mr W last updated on 09/Dec/20

	$x_{n + 1} = {(\frac{81}{x_{n}})}^{\frac{1}{4}}$ $\ln (x_{n + 1}) = \frac{\ln 81}{4} - \frac{\ln (x_{n})}{4}$ $l e t a_{n} = \ln (x_{n}), p = \frac{\ln 81}{4} = \ln 3$ $\Rightarrow a_{n + 1} = p - \frac{a_{n}}{4}$ $l e t a_{n} = b_{n} + c$ $b_{n + 1} + c = p - \frac{b_{n} + c}{4}$ $b_{n + 1} = p - \frac{5 c}{4} - \frac{b_{n}}{4}$ $l e t p - \frac{5 c}{4} = 0 \Rightarrow c = \frac{4 p}{5} = \frac{4 \ln 3}{5}$ $\Rightarrow b_{n + 1} = - \frac{1}{4} b_{n} \Rightarrow G . P .$ $\Rightarrow b_{n} = {(- \frac{1}{4})}^{n} b_{0}$ $b_{0} = a_{0} - c = \ln (x_{0}) - c = \ln 1 - \frac{4 \ln 3}{5} = - \frac{4 \ln 3}{5}$ $a_{n} = {(- \frac{1}{4})}^{n} (- \frac{4 \ln 3}{5}) + \frac{4 \ln 3}{5}$ $a_{n} = [1 - {(- \frac{1}{4})}^{n}] (\frac{4 \ln 3}{5})$ $\ln (x_{n}) = [1 - {(- \frac{1}{4})}^{n}] (\frac{4 \ln 3}{5})$ $\Rightarrow x_{n} = 3^{\frac{4}{5} [1 - {(- \frac{1}{4})}^{n}]} < 3 f o r n ⩾ 2$
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