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Limit of a sequence problem

  1. Jan 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the sequence given by [itex]b_{n} = n - \sqrt{n^{2} + 2n}[/itex]. Taking [itex](1/n) \rightarrow 0[/itex] as given, and using both the Algebraic Limit Theorem and the result in Exercise 2.3.1 (That if [itex](x_n) \rightarrow 0[/itex] show that [itex](\sqrt{x_n}) \rightarrow 0[/itex]), show [itex]\lim b_{n}[/itex] exists and find the value of the limit.

    2. Relevant equations
    [itex]b_{n} = n - \sqrt{n^{2} + 2n}[/itex] and [itex](1/n) \rightarrow 0[/itex]

    3. The attempt at a solution
    Does the [itex](1/n) \rightarrow 0[/itex] imply that I should put bn in the form 1/n? Going in that direction, I'm stuck at [tex]\frac{-2n}{n + \sqrt{n^{2} + 2n}}[/tex] Am I going in the right direction? And if so, any hints on how to further manipulate what I have? I'm self-studying, and do not have a professor or mentor or anything to give me a bit of direction.
     
  2. jcsd
  3. Jan 4, 2016 #2

    Samy_A

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    That certainly seems a right direction. Maybe divide both numerator and denominator by ##n##, and take the ##\frac{1}{n}## into the square root.
     
  4. Jan 4, 2016 #3
    YES! That is the hint I needed! Thank you SO much!!!
     
  5. Jan 4, 2016 #4

    Ray Vickson

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    You can write ##\sqrt{n^2+2n} = \sqrt{n^2} \, \sqrt{1 + (2/n)} = n \sqrt{1 + (2/n)}##. Since you know that ##\lim_{n \to \infty} \sqrt{1+(2/n)} = 1##, it follows that for any (small) number ##r > 0## we have ##1 \leq \sqrt{1+(2/n)} < 1 + r## for all ##n## larger than some number ##N = N_r##. Therefore, for all ##n> N## we have that the denominator of your ratio above lies between ##n + n = 2n## and ##n + n(1+r) = n(2+r)##. Therefore the ratio itself lies between ##-2/2 = -1## and ##-2/2(1+r/2) = -1/(1 + (r/2))##. Therefore, the limit, ##L## is a number that is ## \geq -1## and ##\leq -1/(1+(r/2))##, no matter how small the positive number ##r## that you choose. For example, if ##r = 0.00002## we have ##-1 \leq L <-1/1.00001##. If ##r = .00000000002## then we have ##-1 \leq L < 1/1.00000000001##, etc. Basically, ##L## must be -1; there is no other possible choice.
     
    Last edited: Jan 4, 2016
  6. Jan 5, 2016 #5
    Thank you SO much! I hadn't even thought to use ##\sqrt{n^{2} + 2n} = \sqrt{n^{2}} \sqrt{1 + \frac{2}{n}}##!
    Using my fraction, I somehow got that the limit of ##b_{n}## was -2. I see where my mistake was. When I used [tex]\frac{\frac{-2n}{n}}{\frac{n + \sqrt{n^{2} + 2n}}{n}} = \frac{-2}{1 + \frac{\sqrt{n^2 + 2n}}{n}}[/tex] and then setting the limit as ##n \rightarrow \infty##, I made the mistake of assuming that ##lim_{n \to \infty} \frac{\sqrt{n^{2} + 2n}}{n} = 0## when that is not true.
    Doing it your way, I have ##lim_{n \to \infty} b_{n}## = -1, as do you.

    Thank you so much!!
    The things you learn when you reach out for help!
     
  7. Jan 9, 2016 #6
    You can also do it by induction, much harder and not needed here.
     
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