I have to say I disagree with you.

Consider sequence A1=a, An+1=a^An for a>0.  If the limit x exists, we can take the limit on both sides and will get x=a^x, which is lna=ln(x)/x.  For a is in (0,1), the limit is the solution of lna=ln(x)/x, which is less than 1.

ln(x)/x is 0 at x=1 and x->infinity.  The max value occurs at e^(1/e). 

When a is in (1, e^(1/e)),
lna=ln(x)/x has two solutions.  We need to use the one that is less than e^(1/e) as the limit.

An example is a=2^.5.  (1)An is increase. (2)An<2

Proof of (2)

A1=a<2

Assume An<2 then An+1=a^An <a^2=2.

Therefore An<2 for all n.

Since (1)An is increase. (2)An<2, An has a limit.  Solve ln(2^.5)=ln(x)/x, we get x=2 or x=4. Accept 2 and reject 4 since it is bigger than e^(1/e).

所有跟帖： 

• Thanks,but you are...,Newton was not clear,...100 more years Cau -jinjing- ♀ (287 bytes) () 01/17/2011 postreply 11:12:12

• When I can't your thanks, I check my answer again,my god...... -jinjing- ♀ (846 bytes) () 01/18/2011 postreply 12:31:15

• 回複：When I can't your thanks, I check my answer again,my god..... -yma16- ♂ (548 bytes) () 01/18/2011 postreply 14:22:39

• My friend,you'r great,a>e^(1/e) An is divergent. Why?Because... -jinjing- ♀ (127 bytes) () 01/18/2011 postreply 17:57:16

• thank you for the effort. This is the only -yma16- ♂ (450 bytes) () 01/20/2011 postreply 18:27:29

• Thanks,you are right. A old want to say....why the big one is no -jinjing- ♀ (302 bytes) () 01/21/2011 postreply 10:53:30

• I also found another example that taking a limit -yma16- ♂ (373 bytes) () 01/21/2011 postreply 17:42:54