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Let $x_1 \ge 2, x_{n+1}=1+\sqrt{x_n-1}$. Find the limit, if it is convergent.

My attempt: I'm trying to use the Monotone convergence theorem to show that it is convergent. So, I have to show that the sequence is decreasing and bounded below/ increasing and bounded above.

I have been able to show that it is bounded below by 2 by induction as follows:

$x_1 \geq 2 \Rightarrow x_1-1 \geq 2-1=1$

$\Rightarrow x_2=1+\sqrt{x_1-1}\geq 1+1=2 $.

Let $x_k \geq 2 \Rightarrow x_k-1 \geq 2-1=1$

$\Rightarrow x_{k+1}=1+\sqrt{x_k-1}\geq 1+1=2 $.

How do I show that this is a decreasing sequence?

Then my limit will be given by lim $x_{n+1}$=lim $(1+\sqrt{x_n-1})$

$\Rightarrow l=1+\sqrt {l-1}$

$\Rightarrow l-1=\sqrt {l-1}$

$\Rightarrow (l-1)^2=(l-1)$

$\Rightarrow (l-1)(l-2)=0$

Hence $l=1$ or $2$, but since it is bounded below by $2, l \neq 1$. So limit is $2$.

Please help with the decreasing proof!

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make a change of variable. let $y_n = x_n - 1.$ then $$y_{n+1} = \sqrt y_n,\quad y_1 \ge 1.$$
we have $$1 \le y_1 \implies 1 \le y_1 \le y_1^2=y_2 $$ and by induction, we have $$y_n \le y_{n+1}, n \ge 1. $$

it can also be seen from the fact that $\sqrt x$ has a unique fixed point $x = 1$ and from its graph.

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Note that if $$x_{n+1}<x_n\implies 1+\sqrt {x_n-1}<x_n $$ $$\implies x_n-1<x_n^2-2x_n+1$$ $$\implies x_n^2-3x_n+2>0\implies (x_n-1)(x_n-2)>0$$ which is true since $x_n\geq 2$ for all $n$. Hence, the sequence is a decreasing sequence. Now, for a proof, you just reverse those implications and you can write: Since $x_n\geq 2$ for all $n$, $$(x_n-1)(x_n-2)>0 \implies x_n^2-3x_n+2>0$$ $$\implies x_n-1<x_n^2-2x_n+1\implies x_n-1<(x_n-1)^2$$ $$\implies \sqrt {x_n-1} <x_n-1\implies 1+\sqrt {x_n-1}<x_n \implies x_{n+1}<x_n$$

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I would denote $$y_n:= x_n-1$$

Then you have $y_1 \geq 1$ and

$$y_n=\sqrt{y_{n-1}}$$

You need to show that $y_n \geq 1$ and that $y_n \geq \sqrt{y_n}$, both of which are easy.

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So to show it's decreasing, we need to show that $x_n > x_{n+1}$ (I will be using the conventional $n$). $\Leftarrow x_n > 1+\sqrt{x_n-1} \Leftarrow x_n-1 > \sqrt{x_n-1}$ which is true for all integers > 1. As we have shown above, $\lbrace x_n\rbrace \ge 2$, so this holds.

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  • $\begingroup$ I think you want those arrows to point the other way. $\endgroup$ – AMPerrine May 3 '15 at 3:03
  • $\begingroup$ @AMPerrine can you do an edit as to what you thought? $\endgroup$ – SalmonKiller May 3 '15 at 3:06
  • $\begingroup$ @N.S. The question proves that the sequence is bounded by 2. So the inequality is satisfied for all $x_n$. $\endgroup$ – SalmonKiller May 3 '15 at 3:10
  • $\begingroup$ @N.S. Ok. Thanks. $\endgroup$ – SalmonKiller May 3 '15 at 3:13

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