What Happens to Sqrt(n+1) - Sqrt(n) as n Approaches Infinity?

[tcbh]

Homework Statement

$\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}

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 t_n, moving -\sqrt(n) to the other side and squaring both sides. Then somehow showing that t_n--> 0.

 

[dextercioby]

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.

 

[LeonhardEuler]

Have you tried using the mean value theorem on \sqrt{n+1}? It should come out quickly that way.

 

[tcbh]

Thanks for the help!

I tried multiplying by\sqrt{n}/\sqrt{n}.

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

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 \epsilon so the the expression is always less than \epsilon for n > N. I'm not really sure how to go about constructing that N.

 

[tcbh]

Have you tried using the mean value theorem on \sqrt{n+1}? It should come out quickly that way.

ahh, I see.

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

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

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

 

[dextercioby]

Well, it's still not solved, because you get a new infinity*0 limit.

The next hint is (a+b)(a-b) = a^2 - b^2.

 

[Mark44]

Following along the lines of bigubau's hint, multiply your original expression by 1 in the form of
\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}

 

[LeonhardEuler]

ahh, I see.

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

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

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
0\leq\sqrt{n+1} - \sqrt{n}\leq\epsilon \ \ \ \forall n&gt;\frac{1}{4\epsilon^2}