Show by taking the logarithm (sequences)

Question

I am confused as to how I can show that the sequence

An = (ln(lnn))^(-lnn) is equal to An = n^(-ln(ln(lnn)))

and show that ln(ln(lnn)) >/= 2 if n > C where C = e^(e)^(e)^2

Any hints would be greatly appreciated.

Answer 1

Hint for the first request.

I will use $x$ instead of $n$ to make less confusion.

$$(\ln(\ln x))^{-\ln x} = e^{-\ln(x) \cdot (\ln(\ln(\ln x)))} = n^{-\ln(\ln(\ln x)))}$$

In which I used the exp transformation property:

$$a^b = e^{b\ln a}$$

Now let's solve the second:

$$\ln(\ln(\ln x)) \geq 2$$

Exponentiating once

$$e^{\ln(\ln(\ln x))} \geq e^2$$

That is

$$\ln(\ln x) \geq e^2$$

Exponentiating again

$$e^{\ln \ln x} \geq e^{e^{2}}$$

But $e^{\ln \ln(x)}$ is nothing but $\ln x$$

so

$$\ln x \geq e^{e^{2}}$$

And finally, exponentiating again:

$$x \geq e^{e^{e^{2}}}$$

So actually there is one more $e$ contrarily to what you wrote.