3
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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? $\endgroup$
    – lulu
    May 27, 2020 at 15:21
  • 3
    $\begingroup$ 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. $\endgroup$
    – lulu
    May 27, 2020 at 15:31

2 Answers 2

6
$\begingroup$

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. $$ a_n = \frac{1}{\frac{1}{a}+(n-1)}\,.$$

$\endgroup$
2
  • $\begingroup$ Shouldn't the first expression be $a_n=\frac{1}{b_n}$ ? $\endgroup$
    – VLC
    May 27, 2020 at 16:12
  • 1
    $\begingroup$ I dont think it matters @Bili Debeli $\endgroup$ May 27, 2020 at 17:31
6
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .