How can I find if the sequence $z_n = \sqrt{n+2} - \sqrt{n} $ converges or diverges?

Question

$$z_n = \sqrt{n+2} - \sqrt{n} $$

$$ \lim_{n\to \infty} z_n =\lim_{n\to \infty} \sqrt{n+2} - \sqrt{n} $$

Answer 1

$$\sqrt{n+2}-\sqrt n=\frac 2{\sqrt{n+2}+\sqrt n}$$

Answer 2

HINT $$\sqrt{n+2} - \sqrt{n} = \dfrac2{\sqrt{n+2}+\sqrt{n}}$$

Answer 3

To complement the other answers: $$\sqrt{n+2} - \sqrt{n} = \frac{\sqrt{n+2} + \sqrt{n}}{\sqrt{n+2} + \sqrt{n}}\cdot \sqrt{n+2} - \sqrt{n}$$ $$= \frac{(\sqrt{n+2} - \sqrt{n})(\sqrt{n+2} + \sqrt{n})}{\sqrt{n+2} + \sqrt{n}} = \frac{(n+2) - (n)}{\sqrt{n+2} + \sqrt{n}}$$

Answer 4

Note that $$\left(\sqrt{n}+\frac{1}{\sqrt{n}}\right)^2\gt n+2.$$ It follows that $$\sqrt{n}\lt \sqrt{n+2}\lt \sqrt{n}+\frac{1}{\sqrt{n}},$$ and therefore $$0\lt \sqrt{n+2}-\sqrt{n}\lt \frac{1}{\sqrt{n}}.$$ Now use Squeezing to conclude our limit is $0$.

Answer 5

Nicolas' answer is most beautiful. Let me give some heuristics. Let us calculate $\Big(m+\frac1{100}\Big)^2= m^2 +\frac1{10000}+\frac{2m}{100}$. The last term is unbounded in $m$.

$m^2$ has $m$ as square root. To get a square root that is $1/100$ more than the square root of $m^2$ the quantity you have to add increases without bound in $m$. In your problem we want two numbers differing by fixed number 2 cannot have the difference of square roots constant, it has to keep going down; a sequence of numbers that is decreasing and is positive has to be convergent. Once we get the feeling that it converges, its not difficult to show the limit to be zero.