Can-we-prove-a-a-a-a-a-a-a-gt-1-

Question Number 5081 by FilupSmith last updated on 10/Apr/16

Can we prove :

a + a^{a} + a^{a^{a}} + \dots = \infty, a > 1

Answered by Yozzii last updated on 11/Apr/16

D e f i n e t h e f u n c t i o n f t h a t m a p s

m e m b e r s o f N + {0} t o t h e n - t h s e l f - e x p o n e n t i a t i o n

o f a > 1, s a t i s f y i n g f (n + 1) = a^{f (n)}, f (0) = a . F o r e x a m p l e

f (1) = a^{f (0)} = a^{a} a n d f (2) = a^{f (1)} = a^{a^{a}} .

F u r t h e r d e f i n e t h e s u m S o f t h e f i r s t

n o u t p u t s o f f; i . e S (n) = \underset{i = 1}{\sum^{n}} f (i) o r

S (n) = a + a^{a} + a^{a^{a}} + \dots + a^{⋮^{a} (n - 1 t i m e s s e l f - e x p o n e n t i a t i o n o f a)} .

S u p p o s e f o r c o n t r a d i c t i o n t h a t

\lim_{n \to \infty} S (n) i s f i n i t e . T h e n, \exists N \in N s u c h

t h a t f o r a l l n > N, S (N + 1) = S (N + 2) = \dots = S (n) = \dots .

N o w, S (n) s a t i s f i e s t h e r e c u r r e n c e

e q u a t i o n S (n) - S (n - 1) = f (n) .

S i n c e a > 1, t h e n f (n) > 0 f o r a l l n .

\Rightarrow S (n) - S (n - 1) > 0 o r S (n) > S (n - 1) f o r a l l n \in N .

B u t, b y t h e c o n d i t i o n o f c o n v e r g e n c e

f o r s u f f i c i e n t l y l a r g e n, S (n) = S (n - 1) .

T h i s i s a c o n t r a d i c t i o n t h a t i n d i c a t e s

\lim_{n \to \infty} S (n) i s n o n f i n i t e a n d i n p a r t i c u l a r

S (n) > S (n - 1) s o t h a t S (n) i s d i v e r g e n t .

a + a^{a} + a^{a^{a}} + \dots i s i n f i n i t e l y l a r g e i n

v a l u e .

Commented by FilupSmith last updated on 11/Apr/16

A mazing proof!!