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The task is to find the limit of sequence, given by following recursion: $$ a_1=0, a_{n+1}=\frac{3}{2+a_n} $$ so at first I tried to find some first parts of the sequence $$a_1=0, a_2=\frac{3}{2}, a_3=\frac{6}{7}, a_4=\frac{21}{20}, a_5=\frac{60}{61}, a_6=\frac{183}{182}$$ Maybe I done some kind of mistake in these calculations, but nevertheles It seems like this sequence has a limit equal to 1, and approches it from left for odd indexes $$a_1, a_3, a_5...$$ and from right for even indexes $$a_2, a_4, a_6...$$ Now I had idea to split this sequence to two subsequences, one for odd, and one for even indexes, but one is still connected to another and I got stuck in nowhere. Any clue, how to deal with this task, whitout genereting functions?

Any kind of help would be appriciated.

(PS. I am first year student of maths)

Update: I found that even indexed ones can be described by:

$$a_{2n+2}=\frac{6+3a_{2n}}{7+2a_{2n}} $$

And similary odd one's

$$a_{2n+1}=\frac{6+3a_{2n-1}}{7+2a_{2n-1}} $$

And from that I got Did's hint...(checked).

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    $\begingroup$ You should use the Banach Fixed Point Theorem, applied to $f(x)=3/(x+2)$ on the domain $[0, + \infty)$. $\endgroup$ – Crostul Oct 30 '15 at 10:48
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    $\begingroup$ Hint: Every $a_n$ is nonnegative hence $$|a_{n+1}-1|=\left|\frac{1-a_n}{2+a_n}\right|\leqslant\frac12|a_n-1|.$$ $\endgroup$ – Did Oct 30 '15 at 11:57
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    $\begingroup$ Alternative hint (slightly more advanced): $$\frac{a_{n+1}-1}{a_{n+1}+3}=-\frac13\cdot\frac{a_n-1}{a_n+3}$$ $\endgroup$ – Did Oct 30 '15 at 12:00
  • $\begingroup$ It is unnecessary to split between even and odd indexed terms to use the hints above. $\endgroup$ – Did Oct 30 '15 at 12:42
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Using the advanced hint from Did,

$$\begin{align} \frac{a_n-1}{a_n+3} &=\left(-\frac 13\right)\frac {a_{n-1}-1}{a_{n-1}+3}\\ &=\left(-\frac 13\right)^2\frac {a_{n-2}-1}{a_{n-2}+3}\\ &\vdots\\ &=\left(-\frac 13\right)^{n-1}\frac {a_0-1}{a_0+3}\qquad\qquad\qquad(a_0=0)\\ &=\left(-\frac 13\right)^{n}\underset{\small{\text{as } n\to \infty}}{\longrightarrow} 0\\ a_n-1&\underset{\small{\text{as } n\to \infty}}{\longrightarrow} 0\\ a_n&\underset{\small{\text{as } n\to \infty}}{\longrightarrow} 1\quad\blacksquare \end{align}$$


NB: The advanced hint from Did may be derived by writing the original recurrence relation in the form $$\frac {a_{n+1}}1=\frac 3{2+a_n}=\frac TB$$ and by componendo and dividendo, $$\begin{align}\frac {T-B}{T+3B}:\qquad \qquad \frac {a_{n+1}-1}{a_{n+1}+3}&=\frac {3-(2+a_n)}{3+3(2+a_n)}\\ &=-\frac13\cdot \frac {a_n-1}{a_n+3}\qquad\qquad\qquad\qquad \end{align}$$

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Following @Crostul's hint, consider the function $f(x) = \frac 3{x+2}$ for $x\geqslant 0$. Then for $x,y\geqslant 0$ we have $$|f(x)-f(y)| = 3\left|\frac{x-y}{(x+2)(y+2)}\right| \leqslant \frac34|x-y|, $$ so $f$ is a contraction mapping. It follows from the Banach fixed-point theorem that $f$ has a unique fixed point. By inspection, we see that $$ f(1) = \frac3{1+2}=1. $$ Now, as clearly each $a_n\geqslant 0$, we have $a_{n+1}=f(a_n)$. From the Lipschitz condition we see that $$|f(a_n)-f(1)|=|a_{n+1}-1|\leqslant \frac34|a_n-1|, $$ and hence $$\left|\frac{a_{n+1}-1}{a_n-1} \right|\leqslant \frac34.$$ By induction we may show that $$\left|\frac{a_{n+m}-1}{a_n-1}\right|\leqslant\left(\frac34\right)^m, $$ and hence $$ |a_{n+m}-1|\leqslant\left(\frac34\right)^m|a_n-1|.$$ It follows that $$\lim_{m\to\infty}|a_{n+m}-1|=0.$$ As $n$ is arbitrary, we have $$\lim_{n,m\to\infty}|a_{n+m}-1|=0, $$ from which we conclude that $$\lim_{n\to\infty}a_n=1.$$

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We know that if a sequence $(a_n)$ has a limit $l$, then the sequence $(a_{n+1})$ also has the limit $l$. (You can verify this for your own convenience for any sequence).

Therefore $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_{n}=l$$

So we have $$a_{n+1}=\frac{3}{2+a_n}$$ $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{3}{2+a_n}$$ $$\lim_{n\to\infty}a_{n+1}=\frac{3}{2+\lim_{n\to\infty}a_n}$$ $$l=\frac{3}{2+l}$$ $$l^2+2l-3=0$$ $$(l+3)(l-1)=0$$ $$l=1 \,\ \text{or} \,\ -3$$ But $l$ cannot be negative since $a_1=0$ and $(a_n)$ is a sequence of positive terms oscillating about $1$. So $$\lim_{n\to\infty}a_{n}=l=1$$

EDIT: Take the positive terms $1,1$ and $a_n$. So by A.M-G.M. inequality, we have $$\frac{1+1+a_n}{3}\ge (a_n)^{\frac{1}{3}}$$ $$\frac{3}{2+a_n}\le \frac{1}{(a_n)^{\frac{1}{3}}}$$ $$a_{n+1}\le (\frac{1}{a_n})^{\frac{1}{3}}$$ $$a_{n}\le (\frac{1}{a_{n-1}})^{\frac{1}{3}}$$

Now you can check that $min(a_n)=\frac{6}{7}$ omitting $a_1=0$.

Hence maximum value of $\frac{1}{a_n}=\frac{7}{6}$

So $$a_{n+1}\le (\frac{7}{6})^{\frac{1}{3}}$$

This shows the sequence is bounded.

I don't know but see if you can use this to prove Cauchy.

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    $\begingroup$ Yes, if the sequence converges. But the question is, does it? $\endgroup$ – Did Oct 30 '15 at 11:54
  • $\begingroup$ Yes, that's the point I got stuck. If it converges task is simple, but if it's not, it isn't. I thought about proving that this sequence somehow is a Cauchy sequence, but again I have problem with that, because of its recursion. But I found that I can write odd indexed parts as $$a_{n+2}=\frac{6+3a_n}{7+2a_n} $$ and even indexed ones as sth like $$\frac{7+2a_n}{6+3a_n}$$ Unfortunetly, I am still stuck... $\endgroup$ – Kiwi Oct 30 '15 at 12:00
  • $\begingroup$ @Kiwi I thought you had checked whether the sequence converges or not and hence asked for the limit. Anyway I am checking if the sequence is convergent or not. I'll update my answer as soon as possible. $\endgroup$ – SchrodingersCat Oct 30 '15 at 12:23
  • $\begingroup$ @Aniket I also try to check it by now. Sorry for confusion and thanks for help. $\endgroup$ – Kiwi Oct 30 '15 at 12:25
  • $\begingroup$ @Kiwi I posted two explicit ways to do so in the comments on main, half an hour before your last comment. $\endgroup$ – Did Oct 30 '15 at 12:40
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This is a pure algebraic approach which map the non-linear recurrence relation to a matrix valued linear recurrence relation. Writing $a_n$ as $\displaystyle\;\frac{p_n}{q_n}$, we have:

$$a_{n+1} = \frac{3}{2+a_n}\iff \begin{bmatrix}p\\ q\end{bmatrix}_{n+1} \propto A \begin{bmatrix}p\\ q\end{bmatrix}_n \quad\text{ where }\quad A = \begin{bmatrix}0 & 3\\ 1 & 2\end{bmatrix} $$ Together with $\displaystyle\;a_1 = 0 \implies \begin{bmatrix}p \\ q\end{bmatrix}_1 \propto \begin{bmatrix}0 \\ 1\end{bmatrix}$, this implies $$\begin{bmatrix}p\\ q\end{bmatrix}_n \propto A^{n-1} \begin{bmatrix}0\\ 1\end{bmatrix}$$ Notice the characteristic polynomial for $A$ is $$\det\left(\lambda I_2 - A\right) = \lambda (\lambda - 2) - 3 = (\lambda+1)(\lambda-3)$$

By Cayley-Hamilton theorem, we have

$$(A + I_2)(A - 3I_2) = 0\;\implies\; \begin{cases} A^k (A+I_2 ) &= 3^k (A+I_2)\\ A^k (A- 3I_2)&= (-1)^k(A-3I_2) \end{cases}\quad\text{ for all } k \ge 0$$

Using the partition of identity $$I_2 = \frac14\left((A+I_2) - (A -3I_2)\right),$$ we find

$$A^k = \frac14 A^k\left((A+I_2) - (A -3I_2)\right) = \frac14\left(3^k (A+I_2) - (-1)^k (A-3I_2)\right)$$ and hence

$$\begin{bmatrix}p\\ q\end{bmatrix}_n \propto \left( 3^{n-1}(A+I_2) - (-1)^{n-1} (A-3I_2)\right) \begin{bmatrix}0\\ 1\end{bmatrix} = 3^n \begin{bmatrix}1\\ 1\end{bmatrix} - (-1)^{n-1}\begin{bmatrix}3\\ -1\end{bmatrix}$$ This implies

$$a_n = \frac{p_n}{q_n} = \frac{3^n + 3(-1)^n}{3^n - (-1)^n} = \frac{1 + 3(-\frac13)^n}{1 - (-\frac13)^n} $$ Since both the numerator and denominator converges to $1$ as $n \to \infty$. $a_n$ also converges to $1$.

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