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?

#Algebra #Exponents #Logarithms #Limits

Easy Math Editor

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Comments

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?

Calvin Lin Staff - 6 years 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 $B_n\neq B_{n-1}^n.$

Trevor B. - 6 years ago

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

CS ಠ_ಠ Lee - 6 years ago

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

Jake Lai - 6 years ago

Oh, ok, thanks

CS ಠ_ಠ Lee - 6 years ago

Wait, I think im pretty close.

CS ಠ_ಠ Lee - 6 years 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?

CS ಠ_ಠ Lee - 6 years 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.

amar nath - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$