Sequence based on sequential square root function

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
9 replies · 2K views
cscott
Messages
778
Reaction score
1
Express each term of the sequence [tex]\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...\}[/tex] as a power of 2.

I found [itex]\{2^{\frac{1}{2}}, 2^{\frac{3}{4}}, 2^{\frac{7}{8}}, ...\}[/itex] but I can't get the formula for it so I can find it's limit.
 
Physics news on Phys.org
If I consider just the fractional exponents as a sequence on its own, there is no common difference or common ratio so I'm stuck in this aspect.
 
The numerator and denominator go up by 2, 4, 8, 16, etc., or powers of two.
 
Last edited:
I see... so it'd be like [2^(n) - 1]/2^n ?
 
cscott said:
I see... so it'd be like [2^(n) - 1]/2^n ?
Be careful with the parentheses. It should read:
2 ^ ((2 ^ (n) - 1) / 2 ^ n), or
[tex]2 ^ {\frac{2 ^ {n} - 1}{2 ^ n}}[/tex]
:)
 
Hurkyl said:
Well, I thought he was writing down the formula for the exponent, not the term of his sequence!

Yeah, I meant it only as the exponent.