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Why does the limit(n->∞) sqrt(n)/(sqrt(n)+sqrt(n+1)) equal 1/(1+sqrt(1-1/n))?

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    I am studying for a calculus test tomorrow on this website (http://archives.math.utk.edu/visual.calculus/6/index.html). I am working on the limit comparison test problems but I am unfamiliar with the form they use in their solutions. For example:
    Limit comparison test (prove convergence / divergence)
    Series (from n=1 to ∞) 1/(sqrt(n)+sqrt(n+1))

    2. Relevant equations


    3. The attempt at a solution
    Series (from n=1 to ∞) 1/(sqrt(n)+sqrt(n+1)) compare to Series (from n=1 to ∞) 1/sqrt(n), which we know diverges by the p-series test (p=.5, p<1)
    Let an = 1/(sqrt(n)+sqrt(n+1)) and bn = 1/sqrt(n)
    lim(n->∞) an/bn
    = lim(n->∞) 1/(sqrt(n)+sqrt(n+1)) / 1/sqrt(n)
    = lim(n->∞) sqrt(n)/(sqrt(n)+sqrt(n+1))

    I can see why this limit is 1/2 but I don't understand how the solution to the problem ended up in the form
    lim(n->∞) 1/(1+sqrt(1-1/n)) = 1/2
    I can tell by the other solutions that this form is a rule of sorts but I don't know what it is.
    Thank you for the help.
     
  2. jcsd
  3. Nov 5, 2014 #2

    Mark44

    Staff: Mentor

    Factor ##\sqrt{n}## from the two terms of ##\sqrt{n}## + ##\sqrt{n + 1}##.
     
  4. Nov 5, 2014 #3
    OH. so if you multiply
    sqrt(n)/(sqrt(n)+sqrt(n+1)) by (1/sqrt(n))/(1/sqrt(n)) you get 1/(sqrt(n)+sqrt(n+1))/sqrt(n) which can simplify to 1/(sqrt(n/n)+sqrt(n/n+1/n)) or 1/(1+sqrt(1+1/n))
     
  5. Nov 5, 2014 #4

    Mark44

    Staff: Mentor

    No, it's simpler than that. Just factor ##\sqrt{n}## out of the numerator and denominator. The ##\sqrt{n}## factors can be cancelled, and you're left with the form you see in the solution.
     
  6. Nov 5, 2014 #5
    Oh, ok. I understand. Thank you!
     
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