This is about infinite tetration. see details?

I put up a message asking if the infinite hyper root of 1000 equals 1000^ (1/1000). Some one confirmed that the answer to x^^ infinity is always x^(1/x).

One source claims that the range for x is e-e < x < e1/e

And that is fine, because 3^(1/3), 5^(1/5) etc falls in that range. But this source says that the Y values are only 1/(ee) < y < e(1/e)

That apparently means that (3^(1/3) ^^ infinity is not the cubed root of three.

Even worse this implies that anything above e^(1/e) diverges. So is there no answer to x^^infinity= 50, because 50 is greater than the Y range? How can this be?

This site {http://www.faculty.fairfield.edu/jmac/ther/tower.h... claims this saying "It would seem then that x would equal the cube root of 3 when solving the equation y = 3, but this is not the case. In fact, this procedure will only work when

e-1 < y < e so that 1/(ee) < y < e(1/e)

This has been established by Knoebel [I] and Mitchelmore [2] in their treatments of iterated exponentials. Knoebel introduces the notation nx for these 'hyperpowers' where x, xx, xxx ., . . are

written1x,2x, 3x . . . and y in the example above would be the limit of the sequence y = lim nx.

as n approaches infinity. He then proves that the sequence {nx} converges only in the interval

e-e < x < e1/e

"

can any refute this or confirm this?