For the purpose of this question then the order of operations is what your calculator performs it as.Originally posted by: sao123
I am now thoroughly confused...
My calculator performs a^a^a^a... as a^(a^(a^(...^a))))))...
My calculus book defines a^a^a^a.... as ((((a^a)^a)^a)^a)^......
So which is the correct form of the order of operations?
If we have it the other way then (a^a)^a = a^(2a).
((a^a)^a)^a = (a^(2a))^a = a^(3a) and so on.
(((a^a)^a)^a... n times is equal to a^(na).
Thus taking the limit of n to infinity we have that this value is not finite for any a > 1 and the professor in the original question said that there was an answer greater than 1 where the sum was finite. This only exists when we use the first order of operations you posted.