1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do I solve this limit?

  1. Nov 27, 2012 #1
    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. jcsd
  3. Nov 27, 2012 #2
    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?
     
    Last edited: Nov 27, 2012
  4. Nov 27, 2012 #3
    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
  5. Nov 27, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

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

    RGV
     
  6. Nov 27, 2012 #5
    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?
     
  7. Nov 27, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  8. Nov 27, 2012 #7
    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.
     
  9. Nov 27, 2012 #8
    Nice catch, my bad.
     
  10. Nov 27, 2012 #9
    I'm a sophomore in high school :p
     
    Last edited: Nov 27, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How do I solve this limit?
  1. How do i solve this ? (Replies: 1)

Loading...