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Consider the sequence $\{x_n\}$ defined as $x_1:=2, x_{n+1} = \sqrt{1+x_n}$. It can be shown fairly easily that $\lim\limits_{n\to\infty} x_n = \frac{1+\sqrt{5}}{2}$ by first showing that $x_n$ is a bounded, monotonically decreasing sequence, then passing to the limit of $x_n$. However, I have to demonstrate this using the $\epsilon-N$ definition. By using this thread as a reference, I have attempted a proof. Brief outline of my steps so far:

  1. Show $\{x_n\}$ is bounded and monotone decreasing via induction. Hence, $\{x_n\}$ converges to a limit $L \in \mathbb{R}$.
  2. Note that $\lim\limits_{n\to\infty} x_{n+1} = \lim\limits_{n\to\infty} x_n = L$.
  3. Fix $\epsilon>0$. Then for $n>N_0, N_0 \in \mathbb{N}$ we have $\left| x_{n+1} - L \right| < \epsilon$.
  4. Rewrite as $\left|\frac{\sqrt{1+x_n}-L}{x_n-L}\right|\cdot\left|x_n-L\right|<\epsilon$.
  5. Show that $\left|\frac{\sqrt{1+x_n}-L}{x_n-L}\right| \leq 1 \forall n\in\mathbb{N}$.

Then this is where I'm stuck. I know I should claim that $L:=\frac{1+\sqrt{5}}{2}$, but I'm not sure where to go from there. Any help is appreciated. Thanks.

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    $\begingroup$ I think you may have some slight confusion about how the $\epsilon-N$ definition works. The way to think of it is that you have your sequence $\{x_n\}$ and a potential limit $L$. If you can show that for every $\epsilon > 0$ there exists $N$ such that $|x_n - L| < \epsilon$ for all $n \ge N$, then we can say for sure that $\lim x_n = L$. So step 1 affirms that the limit exists (this is good to do) and makes step 2 valid. Your next step is to show that for every $\epsilon > 0$ there exists $N$ such that $|x_n - \frac{1+\sqrt{5}}{2} | < \epsilon$ for all $n \ge N$. $\endgroup$ – carsandpulsars Jan 12 '18 at 5:33
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$$\left|x_{n+1}-\frac{1+\sqrt5}{2}\right|=\left|\sqrt{1+x_n}-\frac{1+\sqrt5}{2}\right|=\frac{\left|1+x_n-\frac{3+\sqrt5}{2}\right|}{\sqrt{1+x_n}+\frac{1+\sqrt5}{2}}=$$ $$=\frac{\left|x_n-\frac{1+\sqrt5}{2}\right|}{\sqrt{1+x_n}+\frac{1+\sqrt5}{2}}\leq \frac{\left|x_n-\frac{1+\sqrt5}{2}\right|}{1+\frac{1+\sqrt5}{2}}\leq\frac{\left|x_n-\frac{1+\sqrt5}{2}\right|}{2}\leq$$ $$\leq\frac{1}{2^2}\left|x_{n-1}-\frac{1+\sqrt5}{2}\right|\leq...\leq\frac{1}{2^n}\left|x_1-\frac{1+\sqrt5}{2}\right|\rightarrow0.$$ Now easy to use the $\varepsilon-N$ language.

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  • $\begingroup$ Very clear and intuitive proof. Thank you. $\endgroup$ – user513685 Jan 12 '18 at 14:54
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 12 '18 at 14:55
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The missing part from your argument is $$ |x_{n+1}-L|=|\sqrt {1+x_n} - \sqrt {1+L}|=\frac {|x_n-L|}{|\sqrt {1+x_n} +\sqrt {1+L} |} $$ Note that$$ |\sqrt {1+x_n} +\sqrt {1+L} |> 1+ \sqrt {1+L}>2 $$ Therefore $$|x_{n+1}-L|< \frac {|x_{n}-L|}{2} $$The rest of the proof follows easily.

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  • $\begingroup$ I'm confused... if $L=\frac{1+\sqrt{5}}{2}$, then $\sqrt{1+L}<2$. However, you have written the opposite. I do not think that step is valid. $\endgroup$ – user513685 Jan 12 '18 at 14:30
  • $\begingroup$ Thanks, I had missed a 1. It is fixed now. $\endgroup$ – Mohammad Riazi-Kermani Jan 12 '18 at 14:54
  • $\begingroup$ Ah, I see what you're saying now. $\endgroup$ – user513685 Jan 12 '18 at 15:49
  • $\begingroup$ Excellent. Have a good day $\endgroup$ – Mohammad Riazi-Kermani Jan 12 '18 at 15:54

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