More on Sequences — Question 10

*Question 10 Define $a_1=0$ and $a_{n+1}=\dfrac1{2+a_n}$.

1. Show that $0\le a_n\le1/2$ for all $n$.

2. Find the unique fixed point $L$ in this interval and prove that consecutive terms alternate around it.

3. Establish an error contraction and use it to prove $a_n\to L$.

*Question 10 – Solution

The bound follows by induction. The base case $a_1=0$ lies in $[0,1/2]$. If $0\le a_n\le1/2$, then $2+a_n\ge2$, so $0<a_{n+1}=1/(2+a_n)\le1/2$. Hence every term remains in $[0,1/2]$.

A fixed point satisfies $$L=\frac1{2+L}\iff L^2+2L-1=0,$$ so $L=-1\pm\sqrt2$. Only $L=\sqrt2-1$ lies in the invariant interval $[0,1/2]$.

Let $f(x)=1/(2+x)$. Because $f'(x)=-1/(2+x)^2<0$, $f$ is strictly decreasing. Since $a_1<L$, $a_2=f(a_1)>f(L)=L$. Applying the decreasing map reverses the side of $L$ at every iteration, so successive terms alternate across the fixed point.

$$|a_{n+1}-L|=\left|\frac1{2+a_n}-\frac1{2+L}\right| =\frac{|a_n-L|}{(2+a_n)(2+L)}\le\frac14|a_n-L|.$$ Both denominator factors are at least $2$, which justifies the factor $1/4$. Iterating gives $$|a_n-L|\le\left(\frac14\right)^{n-1}|a_1-L|\longrightarrow0.$$ Therefore $a_n\to\sqrt2-1$; the even and odd subsequences approach the same value from opposite sides.