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

  • Thread starter Damidami
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In summary, the limit of the sequence \frac{\sqrt{n+1}}{n} is zero, as shown by the fact that for every \epsilon > 0, there exists n_0 such that for all n \geq n_0, the inequality -\epsilon < \frac{\sqrt{n+1}}{n} < \epsilon holds. This can be achieved by multiplying both sides of the inequality by n, then squaring both sides to create a quadratic inequality, and solving for n. Additionally, the double inequality 0 < \frac{\sqrt{1+n}}{n} <\frac{2}{\sqrt{1+n
  • #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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  • #2
Damidami said:

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
You can use the double inequality

[tex] 0 < \frac{\sqrt{1+n}}{n} <\frac{2}{\sqrt{1+n}} [/tex], valid for n>1.
 
  • #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?
 
  • #5
dextercioby said:
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
Damidami said:
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
Damidami said:
...

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]
 

1. What is the limit of the sequence √(n+1)/n as n approaches infinity?

The limit of the sequence √(n+1)/n as n approaches infinity is 0.

2. How do you calculate the limit of √(n+1)/n as n approaches infinity?

To calculate the limit of √(n+1)/n as n approaches infinity, we can use the limit definition and apply algebraic manipulation to simplify the expression. In this case, we can multiply the numerator and denominator by √(n+1) to obtain the limit of √(n+1)/n as n approaches infinity.

3. What is the significance of this limit in mathematics?

The limit of √(n+1)/n as n approaches infinity is an important concept in the study of sequences and series. It allows us to understand the behavior of a sequence as the number of terms approaches infinity, and is used in various mathematical proofs and applications.

4. Can the limit of √(n+1)/n as n approaches infinity be evaluated using L'Hospital's rule?

Yes, L'Hospital's rule can be used to evaluate the limit of √(n+1)/n as n approaches infinity. By taking the derivative of the numerator and denominator and applying the rule repeatedly, we can simplify the expression and evaluate the limit.

5. Are there any real-life applications of the limit of √(n+1)/n as n approaches infinity?

Yes, the limit of √(n+1)/n as n approaches infinity has various real-life applications, such as in the study of rates of convergence and in the analysis of algorithms. It also has applications in physics and engineering, particularly in understanding the behavior of systems with an infinite number of components.

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