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# 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.
2 years, 5 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 - 2 years, 5 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 $$B_n\neq B_{n-1}^n.$$

- 2 years, 4 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?

- 2 years, 4 months ago

Wait, I think im pretty close.

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Oh, ok, thanks

- 2 years, 4 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.

- 2 years, 4 months ago