Sequence:
$x_1=1$, recursive definition: $x_{n+1}=\sqrt{1+x_n} $ for all n=1

Use induction to prove $x_n$ is bounded above by 2.

I tried:
Bounded above assumption: $x_n \le 2$
Let k=n=1, then $$x_1 \le 2$$ $$1 \le 2$$
Induction:
Let k=n+1, then $$x_{n+1} \le 2$$ $$\sqrt{1+x_n} \le 2$$ $$x_n \le 3$$ And since we assume $x_n \le 2$ holds true for all n, then $2 \le 3$ is true and the sequence is bounded above by 2.

Question

Sequence:
$x_1=1$, recursive definition: $x_{n+1}=\sqrt{1+x_n} $ for all n>=1

Use induction to prove $x_n$ is bounded above by 2.

I tried:
Bounded above assumption: $x_n \le 2$
Let k=n=1, then $$x_1 \le 2$$ $$1 \le 2$$
Induction:
Let k=n+1, then $$x_{n+1} \le 2$$ $$\sqrt{1+x_n} \le 2$$ $$x_n \le 3$$ And since we assume $x_n \le 2$ holds true for all n, then $2 \le 3$ is true and the sequence is bounded above by 2.

perl · Answer

hmmm

perl · Answer

for the induction step, you want to assume that xn is bounded above by 2 , for n = k

perl · Answer

you cant assume that x_n+1 <=2 , that is what you want to prove

zepdrix · Answer

$$\Large\rm x_1=1$$$$\Large\rm x_2=\sqrt{1+1}=\sqrt2$$$$\Large\rm x_3=\sqrt{1+\sqrt2}$$$$\Large\rm x_4=\sqrt{1+\sqrt{1+\sqrt2}}$$$$\Large\rm x_n=\sqrt{1+\sqrt{1+\sqrt{1+...+\sqrt{1+\sqrt{2}}}}}$$Wow this is a weird recursion XD
I'm not even sure if I wrote the x_n correctly lol

perl · Answer

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