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Question Number 61313 by Tony Lin last updated on 31/May/19

$$howtocalculate\underset{n\to +\mathrm{\infty }}{\lim 0}.{05}^{0.{05}^{0.{05}^{{.}^{{.}^{.0.05}}}}}\}n$$$$inasimpleandfastway?$$

Commented by mr W last updated on 31/May/19

$$letx=0.{05}^{0.{05}^{0.{05}^{{.}^{{.}^{.0.05}}}}}\}n\to \mathrm{\infty }$$$$\Rightarrow x=0.{05}^x$$$$\Rightarrow x=e^{x\ln 0.05}$$$$\Rightarrow {xe}^{-x\ln 0.05}=1$$$$\Rightarrow (-x\ln 0.05)e^{-x\ln 0.05}=-\ln 0.05$$$$\Rightarrow -x\ln 0.05=W(-\ln 0.05)$$$$\Rightarrow x=\frac{W(-\ln 0.05)}{(-\ln 0.05)}=\frac{1.0492}{2.9957}=0.3502$$$$i.e.\underset{n\to +\mathrm{\infty }}{\lim 0}.{05}^{0.{05}^{0.{05}^{{.}^{{.}^{.0.05}}}}}\}n\approx 0.3502$$

Commented by mr W last updated on 01/Jun/19

$$generallyfora<1:$$$$a^{a^{a^{.....}}}=\frac{\mathbb{W}(-\ln a)}{(-\ln a)}$$$$fora=1:$$$$a^{a^{a^{.....}}}=1$$$$fora>1:$$$$a^{a^{a^{.....}}}\to \mathrm{\infty }$$

Commented by Tony Lin last updated on 01/Jun/19

$$thankssir,itiscool!$$

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