Infinite Pentation (and x-srt-x)

[tommy1729]

\( \eta < \text{srt}_{k+1}({k+1}) < \text{srt}_k(k) \) for all \( k \ge 5 \)

so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?!

or did i miss something ?

yes, take a similar example:
\( -1 < \frac{1}{k+1} < \frac{1}{k} \)

by your argumentation \( \lim_{k\to\infty} \frac{1}{k} = -1 \).

thats not the same , the limit value cannot possibly be smaller than eta , so eta is the ultimate boundary.

and because any infinite tetration of real base (eta + q) = oo

our resulting limit must be between eta < " limit "< eta + q

taking lim q = 0+ => " limit " = eta.