How do I solve this limit?

  • Thread starter d125q
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  • #1
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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.
 
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Answers and Replies

  • #2
315
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The following limit is to be solved:

$$
\lim_{n\to\infty}(\sqrt{2} \cdot \sqrt[4]{2} \cdot \sqrt[8]{2} \cdot \sqrt[16]{2} \cdot … \cdot \sqrt[2^n]{2})
$$
and I am absolutely clueless about coming up with a solution.

And help will be greatly appreciated!

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

[tex]\prod\limits_{n=1}^{\infty} 2^{\frac{1}{2^n}}[/tex]

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

[tex]2^{1/2}*2^1/4=2^{1/2+1/4}[/tex]

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

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

See where this goes?
 
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  • #3
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As you can see, this:
[tex]\prod\limits_{n=1}^{\infty} 2^{\frac{1}{2^n}}[/tex]
Is equivalent to:
[tex]\sum_{n=1}^\infty 2^{1/2^n}[/tex]

See where this goes?

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!
 
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  • #4
Ray Vickson
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First, I would rewrite this with the pi operator to clean things up:

[tex]\prod\limits_{n=1}^{\infty} 2^{\frac{1}{2^n}}[/tex]

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

[tex]2^{1/2}*2^1/4=2^{1/2+1/4}[/tex]

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

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

See where this goes?

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

RGV
 
  • #5
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No: it is equivalent to
[tex] 2^{\sum_{n=1}^{\infty} 1/2^n}.[/tex]

RGV

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
Ray Vickson
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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?

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

[tex]2^{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 2[/tex]
You're awesome, guys.
 
  • #8
315
2
No: it is equivalent to
[tex] 2^{\sum_{n=1}^{\infty} 1/2^n}.[/tex]

RGV

Nice catch, my bad.
 
  • #9
315
2
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.

I'm a sophomore in high school :p
 
Last edited:

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