Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ [duplicate]
Solution 1:
$f(x)=\frac{1}{\sqrt{x}}$ is a decreasing function on $\mathbb{R}^+$, hence: $$ 2\sqrt{n+1}-2\sqrt{n}= \int_{n}^{n+1}\frac{dx}{\sqrt{x}} \leq \frac{1}{\sqrt{n}} $$ as well as $$ 2\sqrt{n}-2\sqrt{n-1}=\int_{n-1}^{n}\frac{dx}{\sqrt{x}}\geq \frac{1}{\sqrt{n}}.$$
Solution 2:
$$2\sqrt{n+1}-2\sqrt{n}=2\left(\sqrt{n+1}-\sqrt{n}\right)=$$ $$=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}} \Leftrightarrow$$ $$\Leftrightarrow 2\sqrt n<\sqrt{n+1}+\sqrt{n}\Leftrightarrow\sqrt n< \sqrt{n+1}$$