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\lim_{n \to \infty} \sqrt(n+1)/n

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data

    To find the limit of the sequence (I suppose it is zero)
    [tex] \lim \frac{\sqrt{n+1}}{n}[/tex]

    2. Relevant equations
    The limit is zero if for every [itex] \epsilon > 0[/itex] there exists [itex]n_0[/itex] such that for all [itex]n \geq n_0[/itex] one has
    [tex]|a_n - 0| < \epsilon [/tex]
    or what is the same
    [tex]-\epsilon < a_n < +\epsilon[/tex]

    3. The attempt at a solution
    Tried writing it as
    [tex](n+1)^{1/2} n^{-1}[/tex]
    but have no idea what to do next.

    I know I can think the associated differentiable function [tex]f(x) = \frac{\sqrt{x+1}}{x}[/tex] and use L'Hopital, but that would be cheating, as this concept is previous to the limit of functions.
    Last edited by a moderator: Nov 12, 2011
  2. jcsd
  3. Nov 12, 2011 #2


    Staff: Mentor

    You have [itex]\frac{\sqrt{n+1}}{n} < \epsilon[/itex]
    Start by multiplying both sides by n. Since n > 0, this won't change the direction of the inequality. Next, square both sides, which gives you a quadratic inequality that you can solve for n.
  4. Nov 12, 2011 #3


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    You can use the double inequality

    [tex] 0 < \frac{\sqrt{1+n}}{n} <\frac{2}{\sqrt{1+n}} [/tex], valid for n>1.
  5. Nov 12, 2011 #4
    Hi Mark44,
    Thanks, that gives me

    [tex] \alpha_+ = \frac{1 + \sqrt{1+4\epsilon^2}}{2\epsilon^2}[/tex]
    [tex] \alpha_- = \frac{1 - \sqrt{1+4\epsilon^2}}{2\epsilon^2}[/tex]

    So it suffices to take [tex] n_0 = [\alpha_+ + 1] [/tex] right?
    (the least integer greater than the greater of the real roots)

    (It works, still, I think there should be a nicer way, without feching roots of polynomials)

    I have a related question: my teacher used in part of an excercise that
    [tex] \left| \frac{2n - 16}{n^2 + 5} \right| \leq \frac{2n+16}{n^2} [/tex]
    I don't find it obvious, con you find the operation he made to conclude that?
  6. Nov 12, 2011 #5
    Hi dextercioby,
    Thanks! Sounds trivial now, but how did you get that insight?
  7. Nov 12, 2011 #6


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    if n > 1, then n+1 < n+n = 2n.

    but n+1 = (√(n+1))(√(n+1)), for positive n.

    (this is a standard trick with things involving square roots:

    1/√x = √x/x. in this particular instance we use the reciprocals:

    √x = x/√x. so:

    √(n+1)/n = (n+1)/[n√(n+1)] < 2n/[n√(n+1)] = 2/√(n+1).

    the only real insight, besides the "algebraic wizardry"

    is that n > 1 implies n+1 < 2n).
  8. Nov 12, 2011 #7


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    [itex]\displaystyle \left| {2n - 16} \right| \leq {2n+16} [/itex] by using the triangle inequality.

    [itex]\displaystyle n^2 + 5 \ge n^2>0[/itex] pretty obviously.

    Therefore: [itex]\displaystyle \frac{1}{n^2 + 5}\le\frac{1}{n^2}[/itex]
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