# A problem on Cauchy sequences

Let $\langle x_n\rangle$ be a sequence defined recursively by $0<a \le x_1 \le x_2\le b$ and $x_{n+2} =\sqrt{x_nx_{n+1}}$ for each n $\in \Bbb N$ show that $|x_{n+2} -x_{n+1}| \le \frac {b}{b+a} |x_n -x_{n+1}|$ Deduce that $\langle x_n \rangle$ is Cauchy and find its limit.

Can some one help me on this especially in deducing its Cauchy! I think the 1st part can be proven by induction though I had some problem there too.

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To get real angle brackets for the sequences, use \langle ($\langle$) and \rangle ($\rangle$). –  Brian M. Scott Jun 21 '13 at 7:50
To show that $|x_{n+2}-x_{n+1}|\leqslant\cdots$, did you try to write down $|x_{n+2}-x_{n+1}|$ as a function of $x_{n+1}$ and $x_n$? –  Did Jun 21 '13 at 8:01

## 2 Answers

Well, this is not an answer, just an alternative approach.

Let $\theta_n = \ln x_n$. Then we have $\theta_{n+2} = \frac{1}{2} ( \theta_{n+1} + \theta_n)$. It should be clear from this formula that $\theta_n \in [\theta_0, \theta_1]$ for all $n$ (with appropriate adjustments for the relative order of $\theta_0, \theta_1$), so the sequence is well defined.

Then it is easy to compute that $\theta_{n+2}-\theta_{n+1} = \frac{1}{2}(\theta_n - \theta_{n+1})$, which gives $|\theta_{n+2}-\theta_{n+1}| = \frac{1}{2}|\theta_{n+1} - \theta_n|$, and so $|\theta_{n+1} - \theta_n| = \frac{1}{2^n} |\theta_2-\theta_1|$. Then summing gives (assuming $n\ge m$ for simplicity) $$|\theta_{n}-\theta_{m}| =|\theta_{n}-\theta_{n-1}| + ... + |\theta_{m+1}-\theta_{m}| = \frac{2}{2^m} ( 1-\frac{1}{2^{n-m}} )|\theta_2-\theta_1| \le \frac{2}{2^m} |\theta_2-\theta_1|$$

It follows that $\theta_k$ is Cauchy, and since $x \mapsto e^x$ is uniformly continuous on $[\theta_0, \theta_1]$, the sequence $x_k$ is also Cauchy.

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This also makes it easy to compute that $\lim_{n\to \infty} \theta_n = \frac{2\theta_2 + \theta_1}{3}$ (notice OP's indices start at $1$). I love this answer. Ehm... I mean, alternative approach. –  Arthur Jun 21 '13 at 8:21
@Arthur: Very nice observation! –  copper.hat Jun 21 '13 at 8:56

We can estimate \begin{eqnarray} |x_{n+2}-x_{n+1}| & = & |\sqrt{x_{n+1} x_{n+1}}-\sqrt{x_n x_{n+1}}| = \frac{|x_{n+1}(x_{n+1}-x_n)|}{\sqrt{x_{n+1} x_{n+1}} + \sqrt{x_n x_{n+1}}} \\ & = & \frac{1}{1+\sqrt{x_n/x_{n+1}}}|x_{n+1}-x_n| \leq \frac{1}{1+\sqrt{a/b}}|x_{n+1}-x_n| \\ & \leq & \frac{1}{1+a/b/\sqrt{a/b}} |x_{n+1}-x_n| \leq \frac{1}{1+a/b} |x_{n+1}-x_n| \\ & = & \frac{b}{a+b} |x_{n+1}-x_n| \ . \end{eqnarray} Assume $m>n$. Now \begin{eqnarray} |x_m-x_n| & = & \bigg| \sum_{k=n}^{m-1} x_{k+1}-x_k \bigg| \leq \sum_{k=n}^{m-1} |x_{k+1}-x_k| \leq \sum_{k=n}^\infty \bigg(\frac{b}{a+b}\bigg)^{k-1} |x_2-x_1| \\ & = & \frac{1}{1-\frac{b}{a+b}} \bigg(\frac{b}{a+b}\bigg)^{n-1} |x_2-x_1| \rightarrow 0 \ , \end{eqnarray} as $m,n \rightarrow \infty$, where the second inequality is estalished by induction to the index $m-1$. This shows that the sequence is a Cauchy-sequence and has a limit $x$.

The sequence can be mapped using the logarithm function, $y_i = \ln(x_i)$, and we obtain $y_{i+2} = \frac{1}{2}(y_{i+1}+y_i)$ and hence $y_{i+2}-y_{i+1} = -\frac{1}{2}(y_{i+1}-y_i)$. Now \begin{eqnarray} y & = & y_1 + \sum_{k=1}^\infty y_{k+1}-y_k = y_1 + \sum_{k=1}^\infty \bigg(-\frac{1}{2}\bigg)^{k-1} (y_2-y_1) = y_1 + \frac{1}{1+\frac{1}{2}} (y_2-y_1) \\ & = & \frac{y_1+2y_2}{3} \ , \end{eqnarray} where the second equation is established termwise by induction. Hence \begin{eqnarray} x = e^y = e^\frac{y_1+2y_2}{3} = (e^{y_1} e^{y_2^2})^\frac{1}{3}= (x_1 x_2^2)^\frac{1}{3} \ . \end{eqnarray}

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Interesting spelling correction. –  copper.hat Jun 21 '13 at 14:48