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how to prove that the recursive sequence $a_0\ge 0$, $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ is a cauchy sequence? The sequence seems to be bounded and if the sequence is monotonic increasing (I still dont know if it is..), it is convergent, then the sequence must be cauchy. But how to prove with the definition of cauchy sequence if the sequence is cauchy? If I try to start with $|a_{n+1}-a_n|=...$ I don't do useful calculations.. Regards

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Define $f(x) = \dfrac{3 + 3x}{3 + x}, x \geq 0$. It is easily seen that $$f'(x) = \frac{3(3 + x) - (3 + 3x)}{(3 + x)^2} = \frac{6}{(3 + x)^2} \in \left(0, \frac{2}{3}\right]$$ for all $x \geq 0$.

It then follows by the mean value theorem that, for every positive integer $n$: \begin{align*} & \left|a_{n + 1} - a_n\right| = \left|f(a_n) - f(a_{n - 1})\right| \\ \leq & \frac{2}{3}\left|a_n - a_{n - 1}\right| \\ \leq & \cdots \\ \leq & \left(\frac{2}{3}\right)^{n}\left|a_1 - a_0\right| \end{align*} Therefore, for every positive integer $p$, we have \begin{align*} & \left|a_{n + p} - a_n\right| \leq \sum_{k = 1}^p \left|a_{n + k} - a_{n + k - 1}\right| \\ \leq & \sum_{k = 1}^p \left(\frac{2}{3}\right)^{n + k - 1}\left|a_1 - a_0\right| \\ \leq &\left|a_1 - a_0\right|\left(\frac{2}{3}\right)^n\sum_{k = 0}^\infty \left(\frac{2}{3}\right)^k \\ = & 3\left|a_1 - a_0\right|\left(\frac{2}{3}\right)^n \\ \to & 0 \end{align*} as $n \to \infty$, showing that $\{a_n\}$ is a Cauchy sequence.

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  • $\begingroup$ Very Well Done! +1 $\endgroup$ – Mark Viola Aug 9 '15 at 20:48
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Hint.

If you study the function $$f(x)=\frac{3(1+x)}{3+x}$$ you'll see that:

  • $f$ is increasing on $[0,+\infty)$.
  • You have $f(l)=l$ if and only if $l=\sqrt{3}$
  • $f$ is concave on $[0,+\infty)$.

Consequently, you can prove that the sequence $(a_n)$ is:

  • Strictly increasing for $a_0 < \sqrt{3}$ and converging to $\sqrt{3}$.
  • Strictly decreasing for $a_0 > \sqrt{3}$ and converging to $\sqrt{3}$.
  • Constant equal to $\sqrt{3}$ if $a_0=\sqrt{3}$.

In all cases $(a_n)$ is converging (to $\sqrt{3}$) and hence Cauchy.

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  • $\begingroup$ The question explicitly asked how to do it directly using the definition of a Cauchy sequence and not via convergence. $\endgroup$ – joriki Aug 10 '15 at 12:09

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