The limit of the sequence: $U_n=(2\times3)^{-(3\times4)^{..^{\mathrm{-n(n+1)}}}}$

Question

I trying solving this equation here ,I find that the limit l verify: $2=l^{-6^{-12^{..^{\mathrm{-\infty}}}}}$ So now I need the limit of $U_n=(2\times3)^{-(3\times4)^{..^{\mathrm{-n(n+1)}}}}$

Answer 1

$a_n=(1-1/2)^{(1/2-1/3)^{..\mathrm{(1/n-1/(1+n))}}}=(1/2)^{(1/6)^{..\mathrm{1/(n(n+1))}}}=2^{-{6^{-{..^{-{n(n+1)^{\mathrm{-1}}}}}}}}$ suppose this consequence converge to $l$ so it is easy to see that the limit $l$ verify : $2=l^{-{6^{-{..^{-{n(n+1)^{\mathrm{}}}}}}}}$ when n->infinity so $2=l^{-{6^{-{..^{-{\infty^{\mathrm{}}}}}}}}$ because: $a^{b^{\mathrm{-1}}}=l$ require that $a=l^{b}$ @Snaw

Answer 2

I have consider the sequence $U_n=((n-1)n)^{-(n(n+1))}$ this sequence go to $0$ So now : 2=$l^{-U_0^{-..\mathrm{0}}}$ where $U_0=6^{-12}$ and $U_1= 20^{-30}$... until $U_n$