# Can we find a closed-form formula for $a_{n+1}=\frac{a_n}{a_n+1}$, given $a_1=a$?

We have a sequence: $$a_{n+1}=\frac{a_n}{a_n+1}$$ $$a_1 = a$$

$$a_2 = \frac{a}{a+1}$$

$$a_3 = \text{here things get really messy and for all the following }$$ I can't get the formula since the expressions just get more and more complicated. Is there a faster way to do this?

I also have two side questions: What can we say about convergence of this series at the asymptote at $$a_n$$ = -1

The other question that I have is when we test the convergence for other values how much should we prove that $$a_n$$ is increasing/decreasing and how much is enough to just state for instance $$a_{n+1} < a_n$$ without the proof of it.

• I suggest trying it with a few different starting points to get a sense of what happens. If the sequence does converge, to $L$ say, can you see what $L$ has to be?
– lulu
May 27, 2020 at 15:21
• As another suggestion. Define $b_n=\frac 1{a_n}$. Now try a few examples to see what $b_n$ is for various starting points.
– lulu
May 27, 2020 at 15:31

## 2 Answers

If $$a=0$$ then $$a_n=0$$ for all $$n$$.

If $$a_n\neq 0$$ you can write $$b_n = 1/a_n$$. Then you have simply $$b_{n+1} = 1+b_n$$.

It means $$b_n = \frac{1}{a} + n$$ and so $$a_n = \frac{1}{\frac{1}{a} + n}$$.

From this expression you see that if $$a\neq 0$$ you will never have $$a_n=0$$ and that if $$\exists m\in \mathbb{N}$$ such that $$1/a = -m$$ then the sequence is not defined for $$n\geq m$$.

If $$-1/a$$ is not an integer, then the sequence will tend to $$0$$ for $$n\to \infty$$, no matter the value of $$a$$.

EDIT: My answer assumed that $$a_0=a$$. If you want to recover the sequence with $$a_1=a$$, which is what you wrote in your question, you should shift every index, i.e. $$a_n = \frac{1}{\frac{1}{a}+(n-1)}\,.$$

• Shouldn't the first expression be $a_n=\frac{1}{b_n}$ ?
– VLC
May 27, 2020 at 16:12
• I dont think it matters @Bili Debeli May 27, 2020 at 17:31

Hint: If you write the 3rd term for example you start to see the pattern $$a_3 = \frac{\frac{a}{a+1}}{\frac{a}{a+1} + 1} = \frac{a}{2a + 1}$$ and the fourth term is $$a_4 = \frac{\frac{a}{2a+1}}{\frac{a}{2a+1}+1} = \frac{a}{3a + 1}$$ Do you see the pattern? You will need an induction argument to prove this obviously.