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

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 $$1,let $$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?

• $a_2 > 1$ right? So just forget $a_1$ and rename $a_n$ as $b_{n-1}$ Commented Oct 14, 2023 at 5:10
• For $a_n > 0$ we have $1 < a_{n+1}< 2$ Commented Oct 14, 2023 at 8:00

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 $$a_n\to a.$$

• Thanks a lot, your method uses contraction mapping to get the existence of fixed point and avoids the discussion of monotonicity, I totally get it! Commented Oct 14, 2023 at 13:47

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.

• Ah, so I could have just done both $a_{n} \geq a_{n+1}$ and $a_{n} < a_{n+1}$ and the result gives those similar inequalities which shows monotonicity and with the series being bounded; $(a_n)$ converges for $a_{1} > 1$. Commented Nov 12, 2023 at 19:43

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 $$a_{1} > 1.$$

• @MW It is still a massive assumption that all terms of $a_{n}$ are positive, no? Commented Oct 14, 2023 at 6:44
• You are free to write your proofs and responses however you like, and for completeness sake I will do the same for mine. Thanks. Commented Oct 14, 2023 at 6:48
• Fair enough, didn't mean to make this a testy exchange, I'll delete these comments in a sec.
– M W
Commented Oct 14, 2023 at 6:49
• However, it might be worth pointing out that it will not necessarily be true that $$\dfrac{a_{n+1}}{1+a_{n+1}} \geq \dfrac{a_{n}}{1+a_{n}}.$$ We will sometimes have the reverse inequality for all $n$, which is still fine.
– M W
Commented Oct 14, 2023 at 7:06
• @MW Are you referring to my part about not assuming $a_{n}$ is positive? I concede it is clear, but it can still be shown too before the next part using basic facts like you said (and/or induction), so I don't think one must bother with the reverse inequality. Commented Oct 14, 2023 at 7:23