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Question Number 99975 by Dwaipayan Shikari last updated on 24/Jun/20

{(\frac{1}{2})}^{{(\frac{1}{3})}^{\frac{1}{4} \dots . \infty}} = ?

Answered by bachamohamed last updated on 24/Jun/20

s o l u t i o n :

o n a a_{n} = {(\frac{1}{2})}^{{(\frac{1}{3})}^{{(\frac{1}{4})}^{\dots \dots . . (\frac{1}{n + 1})}}} . a_{1} = (\frac{1}{2}) . a_{2} = {(\frac{1}{2})}^{(\frac{1}{3})} = e^{\frac{1}{3} l n \frac{1}{2}}

a_{3} = {(\frac{1}{2})}^{{(\frac{1}{3})}^{(\frac{1}{4})}} = e^{\frac{1}{4} l n ({(\frac{1}{2})}^{(\frac{1}{3})})} = e^{\frac{1}{4} l n (a_{2})} = e^{\frac{1}{4} l n (e^{\frac{1}{3} l n (\frac{1}{2})})} = e^{\frac{2}{4!} l n (\frac{1}{2})}

a_{4} = {(\frac{1}{2})}^{{(\frac{1}{3})}^{{(\frac{1}{4})}^{(\frac{1}{5})}}} = e^{\frac{1}{5} l n ({(\frac{1}{2})}^{{(\frac{1}{3})}^{(\frac{1}{4})}})} = e^{\frac{1}{5} l n (a_{3})} = e^{\frac{1}{5} l n (e^{\frac{2}{4!} l n (\frac{1}{2})})} = e^{\frac{2}{5!} l n (\frac{1}{2})}

a_{5} = {(\frac{1}{2})}^{{(\frac{1}{3})}^{{(\frac{1}{4})}^{{(\frac{1}{5})}^{(\frac{1}{6})}}}} = e^{\frac{1}{6} l n (a_{4})} = e^{\frac{1}{6} l n (e^{\frac{2}{5!} l n (\frac{1}{2})})} = e^{\frac{2}{6!} l n (\frac{1}{2})}

\dots .

\dots

a_{n} = {(\frac{1}{2})}^{{(\frac{1}{3})}^{\dots . . (\frac{1}{n + 1})}} = e^{\frac{1}{(n + 1)} l n (a_{n - 1})} = e^{\frac{1}{(n + 1)} l n (e^{\frac{2}{n!} l n (\frac{1}{2})})} = e^{\frac{2}{(n + 1)!} l n (\frac{1}{2})}

e t : n > 0 (n + 1) < (n + 1)! < {(n + 1)}^{(n + 1)} \Rightarrow n > 0 \frac{2}{(n + 1)} > \frac{2}{(n + 1)!} > \frac{2}{{(n + 1)}^{(n + 1)}}

n > 0 {(\frac{1}{2})}^{(\frac{2}{n + 1})} < {(\frac{1}{2})}^{(\frac{2}{(n + 1)!})} < {(\frac{1}{2})}^{\frac{2}{{(n + 1)}^{(n + 1)}}} \Rightarrow n > 0 . {(\frac{1}{2})}^{(\frac{2}{n + 1})} < a_{n} = e^{\frac{2}{(n + 1)!} l n (\frac{1}{2})} < {(\frac{1}{2})}^{\frac{2}{{(n + 1)}^{(n + 1)}}}

Double subscripts: use braces to clarify

Double subscripts: use braces to clarify

Commented by Dwaipayan Shikari last updated on 24/Jun/20

I h a d a t h o u g h t l i k e y o u . O u r t h i n k i n g i s m a t c h e d . G r e a t s i r

Commented by Dwaipayan Shikari last updated on 24/Jun/20

Just matched!! But i was not so sure about answer . Sir you create my belief

Commented by bachamohamed last updated on 24/Jun/20

{t h a n k}^{'} s