Question

What does (1/2)^(2/3)^(3/4)^(4/5)...^(n/n+1) converge to?

Answer 1

Is that an exponent for like 10 times? Or are those separate ?

Answer 2

Infinite exponentation, somewhat like tetration

Answer 3

Take log, then it is
lim log((1/2)^(2/3)^(3/4)^(4/5)...^(n/n+1) )
=lim log(1/n+1)
=zero
e^0
=1
Therefore the limit is 1.

Answer 4

no, that's not correct, if you evaluate the exponentation out to 15 or so it converges almost immediately on .6044...I want an exact answer.

Answer 5

In multiplication, it is telescopic.
Take log, then it is
lim log((1/2)*(2/3)*(3/4)*(4/5)...*(n/n+1) )
=lim log(1/n+1)
=zero e^0
=1
Therefore the limit is 1 when n->inf.

Answer 6

you're not taking the limit though. think of it like an infinite sum or product, except exponentation.

Answer 7

Does n to to infinity or is it finite?

Answer 8

n goes to infinity. lim does not equal 1, try doing (1/2)^(2/3)^(3/4)^(4/5)^(5/6)^(6/7)^(7/8)^(8/9)^(9/10) and see that it almost immediately converges on .6044

Answer 9

Usually convergences are integral or summation, but for this case it's neither...?

Answer 10

why yes, you are correct.

Answer 11

there are also infinite products, but this is an infinite exponentation.

Answer 12

No idea. I'm used to seeing convergence problems that use integrals and summations, like taylor series or maclaurin series or P series... I can try thinking about it and see if I get anything. Is this calc 2?

Answer 13

You're right, 1 is not the answer. It's a little more complicated.
Will be back if I have new ideas.