# How do I solve this limit?

1. Nov 27, 2012

### d125q

The following limit is to be solved:

$$\lim_{n\to\infty}(\sqrt{2} \cdot \sqrt[4]{2} \cdot \sqrt[8]{2} \cdot \sqrt[16]{2} \cdot … \cdot \sqrt[2^n]{2})$$
all of my attempts to (properly) solve it so far have been futile.

And help will be greatly appreciated!

EDIT: Perhaps this should go to precalc? It's the first time I'm posting here-- so please go easy on me.

Last edited: Nov 27, 2012
2. Nov 27, 2012

### pierce15

First, I would rewrite this with the pi operator to clean things up:

$$\prod\limits_{n=1}^{\infty} 2^{\frac{1}{2^n}}$$

Let's see when you multiply the first two together:

$$2^{1/2}*2^1/4=2^{1/2+1/4}$$

If you multiply the next one:
$$2^{1/2+1/4}*2^{1/8}=2^{1/2+1/4+1/8}$$

As you can see, this:
$$\prod\limits_{n=1}^{\infty} 2^{\frac{1}{2^n}}$$
Is equivalent to:
$$\sum_{n=1}^\infty 2^{1/2^n}$$

See where this goes?

Last edited: Nov 27, 2012
3. Nov 27, 2012

### d125q

If logic serves me well, 1 / 2^n approaches zero, so the whole product equals 1. Is that right (sorry, but I'm rather dizzy right now)?

Thanks a million!

Last edited: Nov 27, 2012
4. Nov 27, 2012

### Ray Vickson

No: it is equivalent to
$$2^{\sum_{n=1}^{\infty} 1/2^n}.$$

RGV

5. Nov 27, 2012

### d125q

Indeed. I am still in high school, so we kind of shun the sigma and pi operators, but they surely make the whole deal a lot easier.

So, does my above post stand?

6. Nov 27, 2012

### Ray Vickson

Parts of it stand, yes, but your final expression is incorrect, as I have plainly indicated. Your sum is infinite, but the correct answer is finite (and not very large).

RGV

7. Nov 27, 2012

### d125q

Oh, damn. I must be blind. So,
$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$
in which case,

$$2^{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 2$$
You're awesome, guys.

8. Nov 27, 2012