Sqrt(n+1) - sqrt(n) as n->oo

  • Thread starter tcbh
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  • #1
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Homework Statement


[tex]$\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}[/tex]


Homework Equations


none

The Attempt at a Solution


The answer is obvious, but I'm having trouble doing it formally. The best I can come up with is setting it equal to some tn, moving [tex]-\sqrt(n)[/tex] to the other side and squaring both sides. Then somehow showing that tn--> 0.
 

Answers and Replies

  • #2
dextercioby
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There's a classic hint on this type of problems. To multiply by 1/1 written in a smart manner, that is involving radicals and <n>. Then taking the limit would be a formality.

If what I said is not obvious, I'll give you a new hint.
 
  • #3
LeonhardEuler
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Have you tried using the mean value theorem on [itex]\sqrt{n+1}[/itex]? It should come out quickly that way.
 
  • #4
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Thanks for the help!

I tried multiplying by[tex]\sqrt{n}/\sqrt{n}[/tex].

Eventually I ended up with [tex]n(\sqrt{1+1/n}-1)/\sqrt{n}[/tex]

Again, it seems pretty clear that the top goes to 0, but I feel like I should be able construct a N based on some [tex]\epsilon[/tex] so the the expression is always less than [tex]\epsilon[/tex] for n > N. I'm not really sure how to go about constructing that N.
 
  • #5
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Have you tried using the mean value theorem on [itex]\sqrt{n+1}[/itex]? It should come out quickly that way.

ahh, I see.

f'(x) = [tex]\sqrt{n+1} - \sqrt{n}[/tex] for some x[tex]\in[/tex](n, n+1)
f'(x) = 1[tex]/2\sqrt{x}[/tex]

Then for x[tex]\geq1/4\epsilon^{2}[/tex], 0<f"(x)<[tex]\epsilon[/tex]

Does that look right (well the tex seems to be a mess, lol)?
 
  • #6
dextercioby
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Well, it's still not solved, because you get a new infinity*0 limit.

The next hint is [itex] (a+b)(a-b) = a^2 - b^2 [/itex].
 
  • #7
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Following along the lines of bigubau's hint, multiply your original expression by 1 in the form of
[tex]\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}[/tex]
 
  • #8
LeonhardEuler
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ahh, I see.

f'(x) = [tex]\sqrt{n+1} - \sqrt{n}[/tex] for some x[tex]\in[/tex](n, n+1)
f'(x) = 1[tex]/2\sqrt{x}[/tex]

Then for x[tex]\geq1/4\epsilon^{2}[/tex], 0<f"(x)<[tex]\epsilon[/tex]

Does that look right (well the tex seems to be a mess, lol)?

This looks right, you just need to use n instead of x technically, so it would be
[tex]0\leq\sqrt{n+1} - \sqrt{n}\leq\epsilon \ \ \ \forall n>\frac{1}{4\epsilon^2}[/tex]
 

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