0
$\begingroup$

I need to show that the Sequence defined by $a_1=1, a_{n+1}=3-\frac{1}{a_n}$ is increasing and $a_n<3$ for all $n$.

I just kind of feel lost when it comes to "showing" a sequence will be increasing, decreasing, or monatomic. Any tips on going about this? Would be greatly appreciated!

$\endgroup$
1
  • $\begingroup$ Try looking at the first few elements of the sequence to get a feeling how the sequence "acts", and then prove by induction. $\endgroup$
    – Itay4
    Mar 6, 2017 at 5:29

2 Answers 2

1
$\begingroup$

$a_{n+1} - a_n = \dfrac{a_n-a_{n-1}}{a_na_{n-1}}$.

Use this to prove that your sequence is increasing. Then the fact that $a_n<3$ follows immediately, because all the terms are positive.

$\endgroup$
2
  • $\begingroup$ @mrnovice Nope, look here $\endgroup$
    – Itay4
    Mar 6, 2017 at 5:46
  • $\begingroup$ It was directed to mrnovice. $\endgroup$
    – Itay4
    Mar 6, 2017 at 8:32
0
$\begingroup$

The following is a stronger statement, which implies the posted one since $1 \gt \frac{3-\sqrt{5}}{2}\,$, $\frac{3+\sqrt{5}}{2}\lt 3\,$.

If $\,a_1 \in (\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})\,$ then the sequence is increasing and $\,a_n \in (\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})\,$ for $\forall n \ge 1\,$.

Note that $x_{1,2}=\cfrac{3\pm\sqrt{5}}{2}$ are the roots of the quadratic $P(x)=x^2-3x+1\,$.

1. $\;\;a_n \gt \cfrac{3-\sqrt{5}}{2}$ $\implies$ $\quad a_{n+1}=3 - \cfrac{1}{a_n} > 3 - \cfrac{2}{3-\sqrt{5}}=3 - \cfrac{2(3+\sqrt{5})}{3^2-5}=3-\cfrac{3+\sqrt{5}}{2} = \cfrac{3-\sqrt{5}}{2}$

2.   Similarly, $a_n \lt \cfrac{3+\sqrt{5}}{2}$ $\implies$ $a_{n+1} \lt \cfrac{3+\sqrt{5}}{2}$

3. $\;\;a_{n+1}-a_n = 3-\cfrac{1}{a_n}-a_n=-\cfrac{a_n^2-3a_n+1}{a_n} \gt 0\,$ because $\,P(x) \lt 0$ for $x \in (x_1,x_2)\,$.

1. and 2. prove that $a_{n+1} \in (\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})\,$, and 3. proves that $a_{n+1} \gt a_n\,$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .