# 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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook