Prove that the limit of $\sqrt{n+1}-\sqrt{n}$ is zero

How would I go about proving that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$? I have tried to use Squeeze theorem but have not been able to come up with bounds that converge to zero. Additionally, I don't think that converting to polar is possible here.

Solution 1:

$$ \sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}} $$

Solution 2:

Using conjugate multiplication can be quite useful in cases like that: $$\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n}){\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}}={(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})\over \sqrt{n+1}+\sqrt{n}}={(n+1)-n\over \sqrt{n+1}+\sqrt{n}}={1\over \sqrt{n+1}+\sqrt{n}}$$. The last result is pretty easy to work with.

Solution 3:

One way is by using the mean value theorem. Specifically, let $f(x) = \sqrt x$. Then, for each $x > 0$, we know that $\displaystyle f(x+1) - f(x) = \frac{f(x+1) - f(x)}{(x+1) - x} = f'(c)$ for some $c$ in the interval $(x, x+1)$. Since $\displaystyle f'(x) = \frac1{2\sqrt x}$ is strictly decreasing we conclude that $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.

Solution 4:

Note that for positive $n$ we have $$\sqrt{n}\lt \sqrt{n+1}\lt \sqrt{n}+\frac{1}{2\sqrt{n}}.\tag{1}$$ The second inequality in (1) holds because $$\left(\sqrt{n}+\frac{1}{2\sqrt{n}}\right)^2=n+1+\frac{1}{4n}\gt n+1.$$ It follows from (1) that $$0\lt \sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}.$$ Now Squeeze.