The infinite tetration of ${x}$ basically means that we take a value and we continue to raise the value to its power forever. If that sounds confusing, it can be thought of as infinite repeated exponentiation of ${x}$. Now, my question is, will this value converge for any vale of ${x}$? In other words, is there a value that this will get infinitely close to when ${x}$ is a certain value?
I got this problem from my teacher as a optional challenge. I am open about this being a given problem, however it is not homework. The problem is stated as follows. Assume we have an infinite tetration as follows $$x^{x^{x^{.^{.^.}}}} \, = \, a$$ With a given $a$ find $x$. The next part of the problem was to discuss the convergence radius of a. If a is too big or too small the tetration does not converge. Below is my humble stab at the problem.  My friend said you would have to treat the tetration as a infinite series, and therefore could not perform algebraic manipulations on it before it is know whether it converges or diverges. However my attempt is to first do some algebraic steps, then discuss the convergence radius. I) Initial discussion At the start it is obvious that the tetration converges when $a=1$ (just set $x=1$) Now after some computer hardwork it seems that the tetration fails to converge when a is roughly larger than 3. II) Algebraic manipulation $$ x^{x^{x^{.^{.^.}}}} \, = \, a$$ This is the same as $$ x^a \, = \, a$$ $$ \log_a(x^a) \, = \, \log_a(a)$$ $$ \log_a(x) \, = \, \frac{1}{a}$$ $$ x \, = \, a^{\frac{1}{a}}$$ Now, if we let $a=2$ then $x = \sqrt{2}$. After some more computational work, this seems to be correct, which makes me believe this formula is correct. III) Discsussion about convergence By looking at the derivative of $ \displaystyle \large a^{\frac{1}{a}} $ we see that the maxima occurs when $a=e$, which also seems to correspond to the inital computational work. Now I think, that the minima of $\displaystyle \large a^{1/a}$ is zero by looking at its graph, studying its derivative and the endpoints. So that my &quot;guess&quot; or work shows that it converges when $$ a \in [0 \, , \,  1/e] $$ VI) My questions Can my algebraic manipulations be justified? They seem rather sketchy taking the a`th logarithm and so on . (Although they seem to &quot;magically&quot; give out the right answer) By looking at Wikipedia it seems that the tetration converge when $$ a \in \left[ 1/e \, , \, e \right] $$ This is almost what I have, why is my lower bound wrong? How can  I find the correct lower bound?
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