If $a_1>1, a_{n+1} = 1+ \frac{a_n}{1+a_n}$, does $a_{n}$ converge?

Question

When $a_1 = 1$, then $a_n$ converges to $\frac{1+\sqrt5}{2}$, But if we let $a_1 > 1$, does the conclusion still hold? Here's what I tried:

First, we can easily conclude $1<a_n<2$,let $f(x) = 1+ x/(1+x),\; (x>1)$, then $f'(x) = 1/(1+x)^2$. So $f(x)$ is strictly increasing. So $$ a_{n+1}> a_n\Leftrightarrow f(a_n) > f(a_{n-1})\Leftrightarrow a_n >a_{n-1} $$ Finally we can get $a_{n+1}> a_n\Leftrightarrow a_2 >a_1$.

Then $$ a_2 - a_1 = 1+a_1/(1+a_1) - a_1 = \dfrac{1+a_1 - a_1 ^2}{1+a_1} $$ The condition is $a_1> 1$, let $g(x)=1+x-x^2\; (x>1)$, the graph of $g(x)$:

Here is the problem : must $a_2 - a_1$ be greater than 0, equals to 0, or less than 0 ?

So is the conclusion incorrect when a is greater than 1?

Answer 1

For a number $a$ we have $$a_{n+1}-a={(2-a)(a_n-a)-(a^2-a-1)\over 1+a_n}$$ If $a={1+\sqrt{5}\over 2}$ we get $$a_{n+1}-a=(2-a){a_n-a\over 1+a_n}$$ Hence $$|a_{n+1}-a|\le(2-a)|a_n-a|$$ Thus $$|a_{n+1}-a|\le (2-a)^n|a_1-a|$$ As $0<2-a={3-\sqrt{5}\over 2}<1$ we get $a_n\to a$ and the speed of convergence is exponential. Hence the series $\sum (a_n-a)$ is convergent.

Remark The above is a general method based on the contraction principle: if $a_{n+1}=f(a_n),$ $f(a)=a$ and $|f'(x)|\le c<1$ then $a_n\to a.$

Answer 2

The only thing you need to know is that the limit exists, and lies in $[1,2]$. Once you know that, letting $L=\lim_{n\to\infty} a_n$, you have $L=1+\frac{L}{L+1}$, giving you the only solution in $[1,2]$ as $L=\frac{1+\sqrt{5}}{2}$.

To show the limit exists, simply observe that every term other than the first is in $[1,2]$, so $(a_n)$ is bounded, and since you have shown $(a_n)$ is monotone (i.e., your same argument shows that if $a_1>a_2$ we get $a_n>a_{n+1}$ for all $n$, so in fact $(a_n)$ is either strictly increasing, strictly decreasing, or constant), we see that it must converge.

Answer 3

For boundedness, $a_{1}>1 \implies 0 \leq \frac{a_n}{1+a_n} < 1 \implies 1 \leq 1+ \frac{a_n}{1+a_n} < 2$.

To show montonocity for $a_{1} > 1$,

Start from $a_{n} \geq a_{n+1}$ and $a_{n} < a_{n+1}$ where the base cases are $a_{1} \geq a_{2}$ and $a_{1} < a_{2}$, respectively.

The result gives inequalities that imply the montonicity of $(a_{n})$, and with the fact that $(a_n)$ is bounded by $[1,2]$, $(a_n)$ converges for all $a_{1} > 1.$