A sequence is defined by $x_1=1$, $x_{n+1}= \sqrt{8+x_n}$. Prove each term of the sequence after the first is greater than the preceding term.

Question

I tried to use mathematical induction, but letting $n=k$ and $n=k+1$ is getting nowhere. I also tried squaring both sides to get rid of the square root but that doesn't seem to help either. Need some hints...

After reading some hints, For $n=k$, assume $x_k > x_{k-1}$

$\sqrt{8+x_{k-1}} > \sqrt{8+x_{k-2}}$

Prove it is true for $k+1$,

$x_k = \sqrt{8+x_{k-1}}$

Squaring both sides,

$8+x_k > 8+x_{k-1}$

$x_k > x_{k-1}$

Since it is true for $n=k+1$, it is also true for $n=k$.

Seems right?

Please show your previous attempts and use the mathjax basic tutorial. — Jessie, Mar 07 '21 at 17:51
If you try to prove the sequence converges, here is the general solution. — rtybase, Mar 07 '21 at 19:40

score 2 · Answer 1 · answered Mar 07 '21 at 17:58

2

Hint

You can try induction. If you suppose that $x_{n}>x_{n-1}$, then

$$8+x_n>8+x_{n-1}\to \sqrt{8+x_{n}}>\sqrt{8+x_{n-1}}.$$

Can you finish?

answered Mar 07 '21 at 17:58

Arnaldo

21,342

Thanks. Please see my edit. Seems right? – Samuel Wong Mar 07 '21 at 18:20
1

@SamuelWong: No, it is not. Take a look at my hint. What is the meaning of the last inequality? – Arnaldo Mar 07 '21 at 18:35
Your last inequality actually implies $x_n$ is greater than its preceding term which is what we're asked to prove? – Samuel Wong Mar 07 '21 at 20:45
@SamuelWong: Yep! – Arnaldo Mar 08 '21 at 13:14
@SamuelWong: If one of the answers satisfy you, could you please, choose one of them (pressing the check mark)? In doing this, you avoid this question pop up many times in the list of question. – Arnaldo Mar 09 '21 at 23:05

score 1 · Answer 2 · answered Mar 08 '21 at 08:57

$$n=1: \;\; x_n = x_1 =1<3=\sqrt{8+1}=\sqrt{8+x_n}=x_{n+1}$$ S.t. it holds for $n$. $$\begin{aligned}n \mapsto n+1: \;\;\;& x_{n+1} \;<\; x_{n+2} \\[5mm] \iff\;\;& x_{n+1} < \sqrt{8 + x_{n+1}} \\[5mm] \iff\;\;& \sqrt{8+x_n} \;<\; \sqrt{8+\sqrt{8+x_n}}\\[5mm] \iff\;\;& 8 + x_n \;<\; 8 + \sqrt{8 + x_n} \\[5mm] \iff\;\;& x_n \;<\; \sqrt{8+x_n} \;\; \iff \;\; x_n \;<\;x_{n+1}\end{aligned} $$ S.t. it is also true for $n+1$, therefore completing the proof by induction.

A sequence is defined by $x_1=1$, $x_{n+1}= \sqrt{8+x_n}$. Prove each term of the sequence after the first is greater than the preceding term.

2 Answers2