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

  • Thread starter Damidami
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  • #1
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Homework Statement



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


Homework 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]


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.
 
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Answers and Replies

  • #2
33,287
4,990

Homework Statement



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


Homework 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]


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.
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.
 
  • #3
dextercioby
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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.
 
  • #4
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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?
 
  • #5
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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.
Hi dextercioby,
Thanks! Sounds trivial now, but how did you get that insight?
 
  • #6
Deveno
Science Advisor
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Hi dextercioby,
Thanks! Sounds trivial now, but how did you get that insight?
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).
 
  • #7
SammyS
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I have a related question: my teacher used in part of an exercise 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?
[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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