Existence of the limit of the recurrent sequence

Question

I need to calculate the limit of the following sequence:

$$ x_n=1\\ x_{n+1}=\frac{30+x_n}{x_n} $$

If it is proven that the limit exists, I know how to do it:

$$ a=\lim_{n\to\infty}\\ a=\frac{30+a}{a}\\ a=6 $$

(we choose positive number because the elements of $x_n$ are positive; it is almost obvious).

But I don't know how to prove the existence of the limit. If you look at the difference $|x_{n+1} - x_n|$, it will be alternating (i.e. the sign is different depending on parity).

So, that is the question: how to prove it?

Thank you in advance

score 1 · Accepted Answer · answered Oct 28 '19 at 08:39

1

You can get rid of the alternation by pairing the iterations:

$$x_{n+2}=\frac{30+x_{n+1}}{x_{n+1}}=\frac{30+31x_n}{30+x_n}>x_n.$$

Thus the even sequence is growing and bounded above by $31$.

answered Oct 28 '19 at 08:39

score 0 · Answer 2 · answered Oct 28 '19 at 06:46

0

Hint: try to prove it is

monotone, i.e. the difference $x_{n+1}-x_n$ is always positive or always negative
bounded, i.e. exits numbers $m,M$ such that $m\leq x_n\leq M$ for all $n$.

answered Oct 28 '19 at 06:46

nonuser

90,026

I tried to prove it is monotone, but it is not (according to Wolfram Alpha). And, as you can see from the question, the difference is alternating – Ivan Andropov Oct 28 '19 at 06:49
Can you prove that the sequences $(x_{2n})$ and $(x_{2n+1})$ are monotone? @IvanAndropov – Angina Seng Oct 28 '19 at 07:12
Yes, I can. But I don't know what should I do next – Ivan Andropov Oct 28 '19 at 07:15

zwim · Answer 3 · 2019-10-28T08:23:54.810

Here is a presentation of the method of adjacent sequences for a similar equation:

Let $a_{n+1}=\dfrac{10}{a_n}-3 \ \ ; \ \ a_1=10$ the find the limits $\lim\limits_{n \to \infty} a_n$

I would like to show a variant, in fact in this case we can even find a closed form for the sequence, and it doesn't cost much more in term of calculations than the method of adjacent sequences.

First we need to notice that $x_n>0$ for all $n$ since $x_1=1$ and there is no subtraction in the induction formula.

Thus the equation is equivalent to $$x_{n+1}x_n=30+x_n$$

Now remark that $30=5\times 5+5$ so let set $y_n=x_n+5$ to cancel the constant term.

We get $(y_{n+1}-5)(y_n-5)=25+y_n\iff$ $$y_{n+1}y_n=5y_{n+1}+6y_n$$

The trick is now to divide by the product (it is $\neq 0$ since $x_n>0$) to get $\ 1=\dfrac 5{y_n}+\dfrac 6{y_{n+1}}$

So let set $z_n=\dfrac 1{y_n}$ and we have a linear equation $$\begin{cases}z_1=\frac 1{x_1+5}=\frac 16\\6z_{n+1}+5z_n=1\end{cases}$$

This solves classicaly to root $-\frac 56$ with initial condition and we find $z_n=\frac 1{11}\left(1-\left(-\frac 56\right)^n\right)$

Since $z_n\to \frac 1{11}\quad$ we get $\quad x_n\to 11-5=6$.

Existence of the limit of the recurrent sequence

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