04/12/2009, 08:58 AM
So this is my attempt at a proof, and there are a bunch of inequalities, which always confuse me 
, so let me know if there are any mistakes.
Let \( \eta = e^{1/e} \) as usual. I will use k (instead of n) to avoid confusion with \( \eta \).
Lemma 1 (Knoebel). \( (a < b) \) iff \( ({}^{k}a < {}^{k}b) \) for all \( a,b,k > 1 \).
Lemma 2. \( \eta < {}^{x}\eta < e \) for all real \( x > 1 \).
Proof.
The base-\( \eta \) tetrational function is continuous and monotonic?
Lemma 3. \( b^x > x + 1 \) for all positive real \( x \ge 5 \) and real \( b > \eta \).
Proof.
The function \( (x+1)^{1/x} < \eta \) for all \( x \ge 5 \), thus \( \eta^x > x+1 \). If \( b > \eta > 1 \), then \( b^x > \eta^x \), so \( b^x > x+1 \).
Lemma 4 (lower bound). For all integer \( k \ge 3 \), \( \eta < a_k = \text{srt}_k(k) \).
Proof.
Since \( k \ge 3 \), then obviously \( k > e \). Together with lemma (2), this implies that \( {}^{k}\eta < e < k \). Substituting \( k = {}^{k}(a_k) \) (hypothesis), this can be written as \( {}^{k}\eta < {}^{k}(a_k) \) which implies \( \eta < a_k \) by lemma (1).
Lemma 5 (decreasing). For all integer \( k \ge 5 \) and \( a_k = \text{srt}_k(k) \), \( a_{k} > a_{k+1} \).
Proof.
We have \( {}^{k}\left(a_{k}\right) = k \) for all integer \( k \) by definition. It follows that
Theorem. \( \lim_{x\to\infty} \text{srt}_x(x) = \eta \)
Proof.
\( \eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k) \) for all \( k \ge 5 \) by lemma (4) and lemma (5).
... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to \( \eta \).
Is this right?
Andrew Robbins
, so let me know if there are any mistakes. Let \( \eta = e^{1/e} \) as usual. I will use k (instead of n) to avoid confusion with \( \eta \).
Lemma 1 (Knoebel). \( (a < b) \) iff \( ({}^{k}a < {}^{k}b) \) for all \( a,b,k > 1 \).
Lemma 2. \( \eta < {}^{x}\eta < e \) for all real \( x > 1 \).
Proof.
The base-\( \eta \) tetrational function is continuous and monotonic?
Lemma 3. \( b^x > x + 1 \) for all positive real \( x \ge 5 \) and real \( b > \eta \).
Proof.
The function \( (x+1)^{1/x} < \eta \) for all \( x \ge 5 \), thus \( \eta^x > x+1 \). If \( b > \eta > 1 \), then \( b^x > \eta^x \), so \( b^x > x+1 \).
Lemma 4 (lower bound). For all integer \( k \ge 3 \), \( \eta < a_k = \text{srt}_k(k) \).
Proof.
Since \( k \ge 3 \), then obviously \( k > e \). Together with lemma (2), this implies that \( {}^{k}\eta < e < k \). Substituting \( k = {}^{k}(a_k) \) (hypothesis), this can be written as \( {}^{k}\eta < {}^{k}(a_k) \) which implies \( \eta < a_k \) by lemma (1).
Lemma 5 (decreasing). For all integer \( k \ge 5 \) and \( a_k = \text{srt}_k(k) \), \( a_{k} > a_{k+1} \).
Proof.
We have \( {}^{k}\left(a_{k}\right) = k \) for all integer \( k \) by definition. It follows that
\(
\begin{tabular}{rl}
a_{k+1}^{\left({}^{k}\left(a_{k+1}\right)\right)} & = {}^{k+1}\left(a_{k+1}\right) \\
1 + {}^{k}\left(a_{k+1}\right) & < {}^{k+1}\left(a_{k+1}\right) \\
1 + {}^{k}\left(a_{k+1}\right) & < k + 1 \\
{}^{k}\left(a_{k+1}\right) & < k
\end{tabular}
\)
by lemma (3). Thus \( {}^{k}\left(a_{k+1}\right) < {}^{k}\left(a_{k}\right) \) which implies \( a_{k+1} < a_k \) by lemma (1).\begin{tabular}{rl}
a_{k+1}^{\left({}^{k}\left(a_{k+1}\right)\right)} & = {}^{k+1}\left(a_{k+1}\right) \\
1 + {}^{k}\left(a_{k+1}\right) & < {}^{k+1}\left(a_{k+1}\right) \\
1 + {}^{k}\left(a_{k+1}\right) & < k + 1 \\
{}^{k}\left(a_{k+1}\right) & < k
\end{tabular}
\)
Theorem. \( \lim_{x\to\infty} \text{srt}_x(x) = \eta \)
Proof.
\( \eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k) \) for all \( k \ge 5 \) by lemma (4) and lemma (5).
... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to \( \eta \).
Is this right?
Andrew Robbins