How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing? How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing?
My idea:


*

*$n^2<(n+1)^2 /+1$

*$n^2+1<(n+1)^2+1/ \sqrt{}$

*$\sqrt{n^2+1}<\sqrt{(n+1)^2+1}/+n$

*$\sqrt{n^2+1}+n<\sqrt{(n+1)^2+1}+n$


And now I'm stuck since if I add 1 to the both sides, I don't know how to move it from the right side without also moving it from the left side.

You have to prove $$\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{\sqrt{(n+1)^2+1}+n+1}$$
i.e., $$\sqrt{n^2+1}+n< \sqrt{(n+1)^2+1}+n+1$$
i.e., $$\sqrt{n^2+1}< \sqrt{(n+1)^2+1}+1$$
i.e., $$n^2+1< (n+1)^2+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$n^2< (n+1)^2+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <2n+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <n+(\sqrt{(n+1)^2+1})$$
Can you conclude this and fill gaps in between?
