How to prove that this sequence is strictly decreasing

[karseme]

\(\displaystyle \dfrac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}\)

\(\displaystyle \dfrac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}>\dfrac{\sqrt{n^2+2n+2}-(n+1)}{\sqrt{n+4}+2(n+1)}\).

So, I've tried to solve this for n, but had no luck. It seems impossible.

\(\displaystyle (\sqrt{n^2+1}-n)(\sqrt{n+4}+2(n+1))>(\sqrt{n+3}+2n)(\sqrt{n^2+2n+2}-(n+1))\)

But, now I will get a bunch of roots, and I don't know how to get rid of them.

 

[Ishuda]

\(\displaystyle \dfrac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}\)

\(\displaystyle \dfrac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}>\dfrac{\sqrt{n^2+2n+2}-(n+1)}{\sqrt{n+4}+2(n+1)}\).

So, I've tried to solve this for n, but had no luck. It seems impossible.

\(\displaystyle (\sqrt{n^2+1}-n)(\sqrt{n+4}+2(n+1))>(\sqrt{n+3}+2n)(\sqrt{n^2+2n+2}-(n+1))\)

But, now I will get a bunch of roots, and I don't know how to get rid of them.

What if that were a function of x? You could use derivatives and then, where necessary, expand the square root function in a binomial series. For example
\(\displaystyle \sqrt{n^2+1}-n = \frac{1}{2n}\, f(n)\)
where f(n) goes to 1 as n goes to infinity.

 

[karseme]

What if that were a function of x? You could use derivatives and then, where necessary, expand the square root function in a binomial series. For example
\(\displaystyle \sqrt{n^2+1}-n = \frac{1}{2n}\, f(n)\)
where f(n) goes to 1 as n goes to infinity.

First of all, my knowledge about derivations is kind of lacking since we still didn't cover them on our lessons and only knowledge that I have about them is just basic. Not to mention that I kind of forgot a lot . I actually hoped if someone could show me how to do it without derivations.

Anyway, I guess my tutors would expect us to solve everything using only methods that we learnt until now offically on our lessons and derivatives are not amongst them yet.

If I multiplied everything I would only get like 3 terms with roots or something like that and it seems impossible to get rid of the roots. Isn't there another way without using derivatives?

What you did there? What is f(n) equal to?

 

[karseme]

I think I found out much easier way to prove this. Thank you on your help.