jaydfox Wrote:Well, most of real and complex analysis would fall apart if limiting cases were not sufficient to provide proofs!
If you ever looked into an analysis book you would know that there are quite rigorous proofs for convergence. You merely telled something about an \( \epsilon \) that goes rapidly to 0. Neither is clear whether this \( \epsilon \) is inside or outside the parenthesis nor why this would imply the convergence of the sequence. And about what limiting cases do you speak?
A proof could perhaps look like this:
We want to show that the sequence \( t_n:=\log_a^{\circ n}(\exp_b^{\circ n}(y)) \) converges.
By the law \( \log_a(b^wr)=w\cdot\log_a(b) + \log_a( r)=w\left(\log_a(b)+\frac{ \log_a ( r ) }{w}\right) \) we inductively construct a supplemental sequence by
\( r_{n,m}=\log_a(b)+\frac{\log_a ( r_{n,m-1} ) }{\exp_b^{\circ n-m}(y)} \) and \( r_{n,0}=1 \).
This sequence is exactly chosen such that
\( t_n=\log_a^{\circ n-m}\left(\exp_b^{\circ n-m}(y)r_{n,m}) \)
particularly
\( t_n=yr_{n,n} \).
Now it is clear by looking at the derivative of \( \log_a \) that \( \log_a(x+\delta)<\log_a(x)+\delta \) for each \( x>1,\delta>0 \). If we repeatedly apply this to the formula of \( r_n \), while assuming that \( 1<a<b \) and hence \( \log_a(b)>1 \), we get
\( r_{n,m}=\log_a(b)+\frac{\log_a ( r_{n,m-1} ) }{\exp_b^{\circ n-m}(y)}\\
\le \log_a(b)+\log_a(\log_a(b))\left(\frac{1}{\exp_b^{\circ n-m}(y)}+\frac{1}{\exp_b^{\circ n-m}(y)\exp_b^{\circ n-m+1}(y)}+\dots+\frac{1}{\exp_b^{\circ n-m}(y)\dots \exp_b^{n-2}(y)}\right) \)
and further for \( n\ge 1 \)
\( t_n\le y\left(\log_a(b)+(\log_a(\log_a(b))\sum_{k=0}^{n-2}\frac{1}{\prod_{j=0}^k \exp_b^{\circ j}(y)}\right) \)
Now is \( \prod_{j=0}^k \exp_b^{\circ j}(y)\ge y^{k+1} \), because \( b^x\ge x \) for \( b\ge\eta \). But we know that the series \( \sum_{k=0}^\infty \frac{1}{y^{k+1} \) converges for \( y>1 \) and hence is the sequence \( t_n \) bounded from above.
An induction shows that \( r_{n,n} \) is increasing.
We show \( r_{n,m}\ge r_{n-1,m-1} \) (for arbitrary \( n, n\ge m \)) by induction over m.
Induction base: \( m=0 \)
\( r_{n,1}=\log_b(a)\ge 1 =r_{n-1,0) \)
and for the induction step show it for \( m\mapsto m+1 \):
From the assumption follows by monotone increase of \( \log_a \):
\( \log_a ( r_{n,m} ) \ge \log_a(r_{n-1,m-1}) \) then
\( \frac{\log_a ( r_{n,m} ) }{\exp_b^{\circ n-(m+1)}(y)}\ge \frac{\log_a ( r_{n-1,m-1} ) }{\exp_b^{\circ (n-1)-m}(y)} \) which yields \( r_{n,m+1}\ge r_{n-1,m} \), the induction assertion.
So particularely \( t_n=yr_{n,n} \) is increasing and bounded from above (for \( y>0 \)) and so has a limit, given that \( b>a \) and that \( b>\eta \).
Theorem. The sequence \( \left(\log_a^{\circ n}({}^{x+n}b)\right)_n \) converges if \( b>a>1 \), \( b>\eta \) (and \( {}^xb>0 \)).
