Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.

Question

Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.

To be monotone it must be either increasing or decreasing, so:

$a_n \ge a_{n-1}$ or $a_n \le a_{n-1}$

$\sqrt{n+1}-\sqrt{n} \ge? \sqrt{n}-\sqrt{n-1} $

$\sqrt{n+1}+\sqrt{n-1} \ge? \sqrt{n} + \sqrt{n}$

I know that $\sqrt{n+1} \ge \sqrt{n}$ and $\sqrt{n} \ge \sqrt{n-1}$

But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.

Can you help me to figure out the solution?

Answer 1

Note that, for $n\geq 1$, $$ (\sqrt{n+1}+\sqrt{n-1})^2 = n+1+2\sqrt{n^2-1}+n-1 = 2n+2\sqrt{n^2-1}\leq 2n +2n=4n, $$ so that $$ \sqrt{n+1}+\sqrt{n-1} \leq 2\sqrt{n} \quad \iff\quad \sqrt{n+1}- \sqrt n \leq \sqrt n - \sqrt{n-1}. $$

Answer 2

hint

use the conjugate

$$a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{b_n}$$

now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.

Answer 3

$$ \\a_{n+1}\le a_{n}<=> \\\sqrt{n+1}+\sqrt{n-1}\le2\sqrt{n}<=> \\(\sqrt{n+1}+\sqrt{n-1})^2\le(2\sqrt{n})^2<=> \\\sqrt{(n-1)(n+1)}=\sqrt{n^2-1}\le \sqrt{n^2}=n $$