Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Technical question regarding showing sqrt(n+1) - sqrt(n) converges to 0

  1. Oct 1, 2009 #1

    gmn

    User Avatar

    1. The problem statement, all variables and given/known data

    show the sequence sn= (n+1)1/2 - n1/2 converges to zero

    2. Relevant equations



    3. The attempt at a solution

    I don't have that much of a problem showing the limit goes to zero, rationalize the numerator (or whatever it's called) to get (n+1)1/2 - n1/2 = 1/((n+1)1/2 + n1/2). My question is that I show this goes to zero because sn<1/n(1/2) which goes to zero, but my professor provides a solution where he writes sn<1/2(n1/2). I don't understand why the 2 is there. Is saying that sn<1/(n1/2) insufficient or not true?

    Thanks
     
  2. jcsd
  3. Oct 1, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    both are true, and sufficient to show it converges to zero, the 2nd is just a little tighter

    how about this, as
    [tex] (n+1)^{1/2} > n^{1/2} [/tex]
    then
    [tex] \frac{1}{s_n} = n^{1/2} + (n+1)^{1/2} > n^{1/2} + n^{1/2} = 2n^{1/2} [/tex]
    then inverting
    [tex] s_n = \frac{1}{n^{1/2} + (n+1)^{1/2}} < \frac{1}{n^{1/2} + n^{1/2}} = \frac{1}{2n^{1/2}} [/tex]
     
  4. Oct 1, 2009 #3
    [tex] \sqrt{n+1}+\sqrt{n}> 2 \sqrt{n}[/tex].
    taking the inverse on both sides yields
    [tex]\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2 \sqrt{n}}[/tex]
    your professor is just using a smaller upper bound for [tex]s_n[/tex]. professors like to use bounds that are as small as possible:smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook