Sequence based on sequential square root function

1. Feb 14, 2006

cscott

Express each term of the sequence $$\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...\}$$ as a power of 2.

I found $\{2^{\frac{1}{2}}, 2^{\frac{3}{4}}, 2^{\frac{7}{8}}, ...\}$ but I can't get the formula for it so I can find it's limit.

2. Feb 14, 2006

Hurkyl

Staff Emeritus
Try a few more then!

Or, it might be better just to write down a way to compute the next term from the current term.

3. Feb 14, 2006

cscott

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.

4. Feb 14, 2006

Hurkyl

Staff Emeritus
Do you at least recognize the pattern?

5. Feb 14, 2006

cscott

The numerator and denominator go up by 2, 4, 8, 16, etc., or powers of two.

Last edited: Feb 14, 2006
6. Feb 14, 2006

cscott

I see... so it'd be like [2^(n) - 1]/2^n ?

7. Feb 14, 2006

Hurkyl

Staff Emeritus
It seems to work for the first three!

8. Feb 15, 2006

VietDao29

Be careful with the parentheses. It should read:
2 ^ ((2 ^ (n) - 1) / 2 ^ n), or
$$2 ^ {\frac{2 ^ {n} - 1}{2 ^ n}}$$
:)

9. Feb 15, 2006

Hurkyl

Staff Emeritus
Well, I thought he was writing down the formula for the exponent, not the term of his sequence!

10. Feb 15, 2006

cscott

Yeah, I meant it only as the exponent.