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If $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2$ and $\lim_{n \rightarrow \infty} b_{n}=1+\sqrt 2 $ , then find $ \ \ \lim_{n \rightarrow \infty} \left|\frac{b_{n+1}-(1+\sqrt 2)}{b_n-(1+\sqrt 2)}\right| $ .

Answer: I think $ \ \ \lim_{n \rightarrow \infty} \left|\frac{b_{n+1}-(1+\sqrt 2)}{b_n-(1+\sqrt 2)}\right| =1$. So the order of convergence $=1$.

Am I right ? Any help is there ?

Martin Sleziak
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MAS
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2 Answers2

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\begin{align} \lim_{n \rightarrow \infty} \left|\frac{b_{n+1}-(1+\sqrt{2})}{b_{n}-(1+\sqrt{2})} \right| &= \lim_{n \rightarrow \infty} \left| \frac{2+\frac{1}{b_n}-(1+\sqrt{2})}{b_{n}-(1+\sqrt{2})} \right| \\ &= \lim_{n \rightarrow \infty} \frac{1}{b_n}\left| \frac{(1-\sqrt{2})b_n+1}{b_{n}-(1+\sqrt{2})} \right| \\&=\lim_{n \rightarrow \infty} \frac{\sqrt{2}-1}{b_n}\left| \frac{b_n+\frac{1}{1-\sqrt{2}}}{b_{n}-(1+\sqrt{2})} \right| \\&=\lim_{n \rightarrow \infty} \frac{\sqrt{2}-1}{b_n}\left| \frac{b_n-(1+\sqrt{2})}{b_{n}-(1+\sqrt{2})} \right| \\&=\lim_{n \rightarrow \infty} \frac{\sqrt{2}-1}{b_n}\\ &=\frac{\sqrt{2}-1}{\sqrt{2}+1} \end{align}

Siong Thye Goh
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Write $c_n=b_n-(1+\sqrt2)$. Then $$b_{n+1}=2+\frac1{1+\sqrt2+c_n} =2+\frac{1}{1+\sqrt2}\left(1-\frac{c_n}{1+\sqrt2}+O(c_n^2)\right) =1+\sqrt2-\frac{c_n}{(1+\sqrt2)^2}+O(c_n^2)$$ etc.

Angina Seng
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