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

tcbh

## Homework Statement

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

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

Homework Helper
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.

Gold Member
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)?

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

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

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

Gold Member
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
$$0\leq\sqrt{n+1} - \sqrt{n}\leq\epsilon \ \ \ \forall n>\frac{1}{4\epsilon^2}$$