Show that $\frac{1}{2 \sqrt{n+1}}\le \sqrt{n+1} - \sqrt{n} \le \frac{1}{2 \sqrt{n}} $

Question

I actually would just like a bit of assistance in understanding a step in the solution. I'm just finding it a bit hard to see how the teacher jumped from one step to one step, if someone could spell it out for me that would be great. Thanks so much!

$ \sqrt{n+1} - \sqrt{n} = \frac{n+1-n}{\sqrt{n+1} + \sqrt{n} }$

I understand this part, as it is just the result of multiplying and dividing by the conjugate. But then the teacher simply has:

Since $ \sqrt{n+1} > \sqrt{n}$

Then $\frac{1}{2 \sqrt{n+1}}< \sqrt{n+1} - \sqrt{n} < \frac{1}{2 \sqrt{n}} $

I probably should know, but could someone just help me out with how he just jumps from the second last step to get the result? It's not obvious to me.

Answer 1

What happens if you increase the denominator? What can you increase to reach required inequality on left? What would happen if you would have decreased it?

Answer 2

This seems to be answered in comments, let me summarize (in a CW answer) what is written there:

We have $$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}.$$ Thus the original inequality is equivalent to $$ \frac1{2\sqrt{n+1}} \le \frac1{\sqrt{n+1}+\sqrt{n}} \le \frac1{2\sqrt{n}}$$ which, in turn, is equivalent to $$2\sqrt{n} \le \sqrt{n+1}+\sqrt{n} \le 2\sqrt{n+1}.$$ The validity of these inequalities should be quite easy to see.

Answer 3

Another approach to the proof, since Martin Sleziak already answered on the step that puzzles you.

This one is a classic application of mean value theorem.

For a differentiable function $f$ on $[a,b]$, you have that

$$\inf_{[a,b]} f'\leq\frac{f(b)-f(a)}{b-a}\leq \sup_{[a,b]} f'$$

Here, $f(x)=\sqrt{x}$ on interval $[n,n+1]$, and $f'(x)=\frac1{2\sqrt{x}}$, which is decreasing, thus

$$\frac1{2\sqrt{n+1}}\leq \frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\leq\frac1{2\sqrt{n}}$$