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$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.

First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$

One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.

Amad27
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  • Hint: easier to show $a_nā‰¤2$. Inductively, if true for $n$ we try $\frac 1{3-a_n}>2\implies 1>6-2a_n\implies 2a_n>5$ a contradiction. ā€“ lulu Jul 27 '16 at 13:25
  • Prove that $2\ge a_{n}\ne 0$ for all $n\in\mathbb{N}$. Now suppose that $a_{n}\le a_{n-1}$. Then $$a_{n}\le a_{n-1}\iff 3-a_{n}\ge 3-a_{n-1}\iff \dfrac{1}{3-a_{n}}\le \dfrac{1}{3-a_{n-1}}\iff a_{n+1}\le a_{n}$$Where did we need the fact that $a_{n}\le 2$? ā€“  Jul 27 '16 at 13:52

3 Answers3

2

First compute, $a_2=\frac{1}{3-2}=1\leq a_1<3$. Then, notice that $$ a_{n+1}-a_{n}=\frac{1}{3-a_n}-\frac{1}{3-a_{n-1}}=\frac{a_n-a_{n-1}}{(3-a_n)(3-a_{n-1})}\leq 0 $$ which gives you the induction step for a strong induction argument for a dual hypothesis: $a_n< 3$ and $a_{n}\leq a_{n+1}$ for all $n\geq 1$. Note that if you already have $a_n<3$, then $a_{n+1}\leq a_n$ will automatically give $a_{n+1}<3$.

yurnero
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Let us go for the overkill, i.e. to find an explicit formula for $a_n$. We may set $a_n=\frac{p_n}{q_n}$, with $p_1=2,q_1=1$, then deduce from $a_{n+1}=\frac{1}{3-a_n}$ the recurrence relation: $$ w_{n+1}=\begin{pmatrix}p_{n+1} \\ q_{n+1} \end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix}p_{n} \\ q_{n} \end{pmatrix}=M w_m\tag{1}$$ The characteristic polynomial of the matrix $M$ is $x^2-3x+1$, having roots $\frac{3\pm\sqrt{5}}{2}=\left(\frac{1\pm\sqrt{5}}{2}\right)^2$.
By the Cayley-Hamilton theorem it follows that $x^2-3x+1$ is also the characteristic polynomial of the sequences $\{p_n\}_{n\geq 1}$ and $\{q_n\}_{n\geq 1}$, hence: $$ p_n = p_{+}\left(\frac{1+\sqrt{5}}{2}\right)^{2n}+p_{-}\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\\ q_n = q_{+}\left(\frac{1+\sqrt{5}}{2}\right)^{2n}+q_{-}\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\tag{2} $$ and both $p_n$ and $q_n$ are linear combinations of the Fibonacci and Lucas numbers $F_{2n}$ and $L_{2n}$.
Interpolating through the values of $p_0,p_1,q_0,q_1$ we get: $$\boxed{ a_n = \color{red}{\frac{F_{2n-5}}{F_{2n-3}}}}\tag{3} $$ and $\{a_n\}_{n\geq 1}$ is a decreasing sequence since $\{F_n\}_{n\geq 2}$ is an increasing sequence.
Moreover, from $(3)$ it follows that: $$ \lim_{n\to +\infty}a_n = \frac{1}{\varphi^2} = \frac{3-\sqrt{5}}{2} \tag{4} $$ as expected, since if $\{a_n\}_{n\geq 1}$ is converging, its limit has to fulfill $L=\frac{1}{3-L}$, i.e. has to be a root of the previous characteristic polynomial.

Jack D'Aurizio
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0

Approach $\boldsymbol{1}$:

Note that $$ \begin{align} a_{n+1}-a_n &=\frac1{3-a_n}-a_n\\ &=\frac{a_n^2-3a_n+1}{3-a_n}\\ &=\frac{\left(a_n-\frac32\right)^2-\frac54}{3-a_n}\tag{1} \end{align} $$ If $\frac{3-\sqrt5}2\le a_n\le\frac{3+\sqrt5}2$, then $(1)$ is negative and $$ \frac{3-\sqrt5}2\le\frac1{3-a_n}\le\frac{3+\sqrt5}2\tag{2} $$ Furthermore, $\frac{3-\sqrt5}2\le2\le\frac{3+\sqrt5}2$.

Approach $\boldsymbol{2}$:

Note that $\frac1{3-x}$ is an increasing function of $x$ on $(0,3)$. Furthermore, if $a_n\lt3$, then $\frac1{3-a_n}\gt0$.

Since $$ a_1=2\gt a_2=1\tag{3} $$ by repeatedly applying $\frac1{3-x}$ to $(3)$, we get that $$ a_n\gt a_{n+1}\tag{4} $$

robjohn
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