What is $\displaystyle\lim_{n\rightarrow\infty}\dfrac{A}{B},$ where $A=n^{(n-1)^{(n-2)^{\ldots^{3^2}}}}$ and $B=2^{3^{4^{\ldots^{(n-1)^n}}}}$? I suspect it is $0,$ but I don't have any idea how I would go about proving this. A little help?

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## Comments

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TopNewestThose terms should be defined as $A_n$ and $B_n$.

Hint:How does $A_n$ and $A_{n-1}$ relate? What about $B_n$ and $B _{n-1}$?Which test does that suggest we apply?

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While it is obvious that $A_n=n^{A_{n-1}},$ I can't come up with a mathematical relartion for $B_{n-1}$ and $B_n,$ since $B_n\neq B_{n-1}^n.$

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The way I look at it, the limit of $\frac{A}{B}$ would be in the form $\frac{\inf}{\inf}$,

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Certainly, but the question would be which sequence (see Calvin Lin's comment) grows faster.

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Oh, ok, thanks

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Wait, I think im pretty close.

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I have a question, though. If the B grows faster, than the equation would near zero. If A grows faster, the equation would near infinity... right?

Am I missing something right now?

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suppose the value of pie =0.31825. calculate the area of circle of radius= 5 equal to 78.55 . please share the answer ok.

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