2
$\begingroup$

Prove that the sequence given by its general term:

$$\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}$$

is strictly decreasing.

$a_n>a_{n+1} \Leftrightarrow \frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}>\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$

Is there some kind of trick to apply in these sorts of problems or do I have to solve this by sheer algebraic manipulation? If I subtract $\frac{\sqrt{(n+1)^2+1}-(n+1)}{\sqrt{(n+1)+3}+2(n+1)}$ from both sides I'll end up with something really ugly so I'm wondering if there's a better way.

1 Answers 1

6

Hint:

$$a_n=\frac{\sqrt{n^2+1}-n}{\sqrt{n+3}+2n}=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}=\frac1{(\sqrt{n+3}+2n)(\sqrt{n^2+1}+n)}$$

  • 1
    Beat me by 12s... +1 – 2017-01-05
  • 0
    I'm sorry about the late response... I understand that $a_{n+1}=\frac{1}{{(\sqrt{n+4}+2n+2)}{(\sqrt{n^2+2n+2}+n+1})}$ so $a_{n+1}$ will have a greater denominator so it'll be smaller than $a_n$, but do I still need to do some algebra to prove it, or do I leave it like that? – 2017-01-09