User blog:Simply Beautiful Art/Slightly better lower bounds to f 3 in the fast growing hierarchy

See also.

I will be proving the lower bound of

$$f_3(n)\ge n2^n((2^{n2^n})\uparrow\uparrow(n-1))$$

where we have the fast growing hierarchy and tetration.

I will be proving this by induction in the same manner I have proved the other bounds. We start by rewriting the problem as

$$f_2^j(n)\ge n2^n((2^{n2^n})\uparrow\uparrow(j-1))$$

Trivially this holds for $$j=1$$. Assume it holds for some arbitrary $$k\ge1$$. Then, we shall prove the above by induction:

$$\begin{align}f_2^{k+1}(n)&=f_2(f_2^k(n))\\&=f_2^k(n)2^{f_2^k(n)}\\&\ge f_2(n)2^{f_2^k(n)}\\&=n2^n2^{f_2^k(n)}\\&\ge n2^n2^{n2^n((2^{n2^n})\uparrow\uparrow(k-1))}\\&=n2^n(2^{n2^n})^{(2^{n2^n})\uparrow\uparrow(k-1)}\\&=n2^n((2^{n2^n})\uparrow\uparrow k)\end{align}$$

Q.E.D.