<s>[INST] Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge? Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge ? If then find the limit . Also the order of convergence $ \eta $.
Answer: \begin{align} b_{0 }=2 \\ b_{1}=2+\frac{1}{2} \\ b_{2}=2+\frac{1}{2+\frac{1}{2}} \\ b_{3}=2+\frac{1}{2+ \frac{1}{2+\frac{1}{2}}}\\ ..... \end{align} $ Actually \ this \ sequence \ is \ a \ continued \ fraction. $  So the sequence is convergent.
Let $ \lim_{n \rightarrow \infty } b_{n}=b $ , then taking the limit of both sides of $ b_{n}=2+\frac{1}{b_{n-1}} $ , we get
$ a=2+\frac{1}{a} $
or, $ a^{2}-2a-1=0 $ ,
or, $ a=\frac{2 \pm 2 \sqrt 2}{2}=1 \pm \sqrt 2 $ . But $ a \nless 0 $ .
So $ a=1+\sqrt 2=2.414 $.
Am I true ? Any help is there ?
 [/INST] It's obvious that $b_n>2$.
Thus, $$|b_n-1-\sqrt2|=\left|\frac{1}{b_{n-1}}-\frac{1}{1+\sqrt2}\right|=|\frac{|b_{n-1}-1-\sqrt2|}{(1+\sqrt2)b_{n-1}}<\frac{|b_{n-1}-1-\sqrt2|}{2(1+\sqrt2)},$$
which says that $\{b_n\}$ converges to $1+\sqrt2$.
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