1
$\begingroup$

Consider the sequence defined by $a_1 = 1$ and $a_n =\frac{1}{1 + a_{n−1}}$ for all $n ≥ 2$. In general, this sequence can also be described as the sequence of quotients of succesive terms in the Fibonacci sequence. Prove that $(a_n)_{n\in \Bbb N}$ is a Cauchy sequence.

  • 0
    well do you know that every convergent sequence is also cauchy – 2017-02-25
  • 0
    @Jeff Hoffman: Is the point of the exercise to show the sequence is Cauchy from the definition of Cauchy sequence -- or just take the easier route of declaring it to be Cauchy by quoting or proving convergence some other way? – 2017-02-25
  • 0
    @RRL to show the sequence is Cauchy from the definition of Cauchy sequence – 2017-02-25
  • 1
    Show that : |$a_{n+1}-a_n$| $\leq 4/9$|$a_n-a_{n-1}|$ – 2017-02-25
  • 0
    this shows that the difference between consecutive terms is getting smaller ie( cauchy) , try proving this with induction – 2017-02-25
  • 0
    what i said implies what you said. – 2017-02-25
  • 0
    ok! thanks for your help – 2017-02-25

2 Answers 2

2

Do you know that the ratios of terms in the Fibonacci sequence converge to the Golden ratio? And of course, convergent sequences are always Cauchy.

  • 0
    my instructor does not give us info about golden ratio.So I have no idea about how to prove it to be cauchy. – 2017-02-25
  • 0
    Sorry, with your definition, it limit of the ratios is 1/golden ratio. See https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression for a closed form expression of the Fibonacci numbers, which should be enough for you to determine the limit of the ratios. – 2017-02-25
  • 0
    can you give me a quick proof of this question? – 2017-02-25
2

Given the Fibonacci sequence where $a_{n+1} = a_n + a_{n-1}$, we have with $b_n = a_n/a_{n-1},$

$$\begin{align} |b_{n+1} - b_{n}| &= \left|\frac{a_{n+1}}{a_n}- \frac{a_{n}}{a_{n-1}} \right| \\ &= \left|\frac{a_{n+1}a_{n-1} - a_n^2}{a_n a_{n-1}} \right| \\ &= \left|\frac{a_na_{n-1} + a_{n-1.}^2 - a_n a_{n-1} - a_n a_{n-2}}{a_{n-1}^2 + a_{n-1}a_{n-2}} \right|. \end{align}$$

Note that the sequence is increasing and $a_{n-1}^2 + a_{n-1}a_{n-2} > 2 a_{n-1}a_{n-2}.$

Hence,

$$\begin{align} |b_{n+1} - b_{n}| &< \left|\frac{a_{n-1}^2 - a_{n}a_{n-2}}{2 a_{n-1}a_{n-2}}\right| \\ &= \frac{1}{2}\left|\frac{a_n}{a_{n-1}} - \frac{a_{n-1}}{a_{n-2}}\right| \end{align}.$$

By induction we can show that

$$|b_{n+1} - b_{n}| < \left(\frac{1}{2} \right)^{n-2}\left|\frac{a_2}{a_1} - \frac{a_1}{a_0} \right| = \left(\frac{1}{2} \right)^{n-2},$$

and

$$|b_{n+k} - b_{n}| < \sum_{j=0}^{k-1} (1/2)^{n-2 +j} = (1/2)^{n-2} \frac{1 - (1/2)^k}{1 - 1/2} < (1/2)^{n-3}.$$

Since the RHS converges to $0$ as $n \to \infty$ the sequence is Cauchy.