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Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1}-2}$ for $n \in \mathbb{N}$. Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ .

I divided this problem into $3$ cases but I don't know whether it is correct or not if there is anything wrong please tell me? I will explain what I did briefly?

$1^{st}$ case

if $x_0=2$ or $x_0=3$ it constant sequence so that either $\lim_{n \to \infty}x_n=2$ or $3$ .

$2^{nd}$case

if $2<x_0<3$ then I showed $x_n>x_{n-1}$ (increasing sequence) and $x_{n-1}<3$ using induction

$3^{rd}$case

if $x_0>3$ then I showed $x_n<x_{n-1}$ (decreasing sequence) and $x_{n-1}>3$ using induction

then we can say $\lim_{n \to \infty}x_n=2$ or $3$ (Monotone convergence theorem) since in either case $\{x_n\}$ is increasing or decreasing and bounded below or bounded above

In other word in either case, $$L=\lim_{n\to \infty}x_{n+1}=2+\sqrt{L-2}\iff L=2\text{ or }L=3$$

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I don't think you can use monotone convergence directly to show that the limit is $2$ or $3$. As you have done though, you can show that it means for any $x_0 \geq 2$ the limit exists. Once you have this, you can write (using continuity of $\sqrt{.}$) $L = 2 + \sqrt{L-2}$ as you have done, and solve for $L$. So just to summarize:

  1. Use monotone convegence to show the limit exists and is nonnegative for any $x_0$
  2. Use continuity of $\sqrt{.}$ to get $L = 2 + \sqrt{L-2}$.
  3. Conclude $L = 2$ or $L = 3$.
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  • $\begingroup$ why I can't apply monotone convergence? I didn't use it directly $\endgroup$
    – ALMEra
    Commented Dec 6, 2020 at 17:07
  • $\begingroup$ You can apply it, but before your edit you wrote in each case that '$x_{n-1} < 3$' is an increasing sequence and so $x_{n-1} \rightarrow 3$' which is false, you only know it converges to something less than or equal to $3$. You've fixed that in your edit, and so now the solution looks better. You still need to mention you are using continuity of $\sqrt{.}$ though to get the equation involving $L$. $\endgroup$
    – xeroxPark
    Commented Dec 6, 2020 at 17:10

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