# Prove $\sqrt{n^2 + 1} - n$ is strictly decreasing

I need to prove the following:

$$\sqrt{n^2 + 1} - n$$ is decreasing

In other words, prove: $$\sqrt{(n+1)^2 + 1} - (n+1) < \sqrt{n^2 + 1} - n$$

Logically I understand why this holds, since the bigger $$n$$ gets, the closer $$\sqrt{n^2 + 1}$$ gets to $$\sqrt{n^2} = n$$, but I don't know how to prove this algebraically.

hint

Multiply by the conjugate to get

$$\sqrt {n^2+1}-n=\frac {1}{\sqrt {n^2+1}+n}$$

the sequence in the denominator is increasing.

• Best way of course! (+1)
– user
Mar 28, 2018 at 22:15
• (+1) Your way is better than mine. Mar 28, 2018 at 22:25

Thinking of $n$ as a real variable we find that $$\frac{\mathrm{d}}{\mathrm{d}n} \left( \sqrt{n^2+1}-n \right)=\frac{n}{\sqrt{n^2+1}}-1<0.$$

You have\begin{multline}\sqrt{(n+1)^2+1}-(n+1)<\sqrt{n^2+1}-n\iff\\\iff\sqrt{(n+1)^2+1}-\sqrt{n^2+1}<n+1-n=1.\end{multline}But$$\sqrt{(n+1)^2+1}-\sqrt{n^2+1}=\frac{2n+1}{\sqrt{(n+1)^2+1}+\sqrt{n^2+1}}$$and it is true that this is less than $1$, because$$\sqrt{(n+1)^2+1}+\sqrt{n^2+1}>\sqrt{(n+1)^2}+\sqrt{n^2}=2n+1.$$