# Exponentiation Question

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? Note by Trevor B.
5 years, 6 months ago

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Those 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?

Staff - 5 years, 6 months ago

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 $B_n\neq B_{n-1}^n.$

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

Oh, ok, thanks

- 5 years, 6 months ago

Wait, I think im pretty close.

- 5 years, 6 months ago

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?

- 5 years, 6 months ago

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

- 5 years, 6 months ago