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

how to calculate lim_(n→+∞) 0.05^(0.05^(0.05^.^.^(.0.05) ) ) }n in a simple and fast way?

$$\left.{how}\:{to}\:{calculate}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{.^{.^{.\mathrm{0}.\mathrm{05}} } } } } \right\}{n} \\ $$$${in}\:{a}\:{simple}\:{and}\:{fast}\:{way}? \\ $$

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

let x=0.05^(0.05^(0.05^.^.^(.0.05) ) ) } n→∞ ⇒x=0.05^x ⇒x=e^(xln 0.05) ⇒xe^(−xln 0.05) =1 ⇒(−xln 0.05)e^(−xln 0.05) =−ln 0.05 ⇒−xln 0.05=W(−ln 0.05) ⇒x=((W(−ln 0.05))/((−ln 0.05)))=((1.0492)/(2.9957))=0.3502 i.e. lim_(n→+∞) 0.05^(0.05^(0.05^.^.^(.0.05) ) ) }n ≈0.3502

$$\left.{let}\:{x}=\mathrm{0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{.^{.^{.\mathrm{0}.\mathrm{05}} } } } } \right\}\:{n}\rightarrow\infty \\ $$$$\Rightarrow{x}=\mathrm{0}.\mathrm{05}^{{x}} \\ $$$$\Rightarrow{x}={e}^{{x}\mathrm{ln}\:\mathrm{0}.\mathrm{05}} \\ $$$$\Rightarrow{xe}^{−{x}\mathrm{ln}\:\mathrm{0}.\mathrm{05}} =\mathrm{1} \\ $$$$\Rightarrow\left(−{x}\mathrm{ln}\:\mathrm{0}.\mathrm{05}\right){e}^{−{x}\mathrm{ln}\:\mathrm{0}.\mathrm{05}} =−\mathrm{ln}\:\mathrm{0}.\mathrm{05} \\ $$$$\Rightarrow−{x}\mathrm{ln}\:\mathrm{0}.\mathrm{05}={W}\left(−\mathrm{ln}\:\mathrm{0}.\mathrm{05}\right) \\ $$$$\Rightarrow{x}=\frac{{W}\left(−\mathrm{ln}\:\mathrm{0}.\mathrm{05}\right)}{\left(−\mathrm{ln}\:\mathrm{0}.\mathrm{05}\right)}=\frac{\mathrm{1}.\mathrm{0492}}{\mathrm{2}.\mathrm{9957}}=\mathrm{0}.\mathrm{3502} \\ $$$$\left.{i}.{e}.\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{\mathrm{0}.\mathrm{05}^{.^{.^{.\mathrm{0}.\mathrm{05}} } } } } \right\}{n}\:\approx\mathrm{0}.\mathrm{3502} \\ $$

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

generally for a<1: a^a^a^(.....) =((W(−ln a))/((−ln a))) for a=1: a^a^a^(.....) =1 for a>1: a^a^a^(.....) →∞

$${generally}\:{for}\:{a}<\mathrm{1}: \\ $$$${a}^{{a}^{{a}^{.....} } } =\frac{\mathbb{W}\left(−\mathrm{ln}\:{a}\right)}{\left(−\mathrm{ln}\:{a}\right)} \\ $$$${for}\:{a}=\mathrm{1}: \\ $$$${a}^{{a}^{{a}^{.....} } } =\mathrm{1} \\ $$$${for}\:{a}>\mathrm{1}: \\ $$$${a}^{{a}^{{a}^{.....} } } \rightarrow\infty \\ $$

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

thanks sir, it is cool!

$${thanks}\:{sir},\:\:{it}\:{is}\:{cool}! \\ $$

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