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Let $\left(\,x_{n}\,\right)_{\ n\ \geq\ 1}$ and $\left(\,y_{n}\,\right)_{\ n\ \geq\ 1}$ be two sequences defined as follows $\left(\,x_{1},y_{1} > 0\,\right)$: $$ \left\{\begin{array}{rcl} {\displaystyle x_{n + 1}} & {\displaystyle =} & {\displaystyle{1 \over 3}\left(\,x_{n} + 2y_{n}\,\right)} \\[2mm] {\displaystyle y_{n + 1}} & {\displaystyle =} & {\displaystyle{1 \over 3}\left(\,{1 \over x_{n}} + {1 \over 2y_n}\,\right)} \end{array}\right. $$ Show that the sequences are convergent and find their limits.

MY TRY:

  • I think about using the Weierstrass' Theorem ( every bounded and monotone sequence is convergent ), but I can't find the monotonity for any sequence.
  • Afterwards, I can think of denoting the two limits as $L_{1}$ and $L_{2}$ and replacing them in the reccurence relationship.
Felix Marin
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M. Stefan
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  • I can make the sequences disjoint (one relation for $x_n$ only and independent one for $y_n$ only). But I don't know if knowing that $(y_n)$ satisfies $y_{n+1}=\frac{1}{2}+\frac{1}{6y_n}-\frac{1}{12y_ny_{n-1}}$ (this is what I obtained for this sequence) makes the problem easier. – tong_nor Jul 25 '17 at 19:43
  • this is a good idea to get an idea to tackle this problem – Dr. Sonnhard Graubner Jul 25 '17 at 19:44
  • $x_n \rightarrow \color{red}{\frac{1}{\sqrt{2}}}$ and $y_n \rightarrow \color{red}{\frac{1}{\sqrt{2}}}$. Should be easy to prove ... ? – Donald Splutterwit Jul 25 '17 at 21:04
  • $x_{n+1} - x_n = \frac{2}{3}(y_n - x_n)$. Does this help? – A. Thomas Yerger Jul 26 '17 at 16:51
  • This was harder than expected. Where did you get this problem? – Aryabhata Jul 30 '17 at 23:31
  • @aryabhata Romanian Mathematical Gazette – M. Stefan Aug 01 '17 at 12:09
  • @M.Stefan: Thanks! Looks like a nice collection of problems. Can you please point me to the exact problem? I presume it is Analysis -> Limit something? I am curious to know the other solutions. – Aryabhata Aug 01 '17 at 15:02
  • It is a problem from the Mathematical Gazette number 4/2017 - April. – M. Stefan Aug 01 '17 at 15:14
  • @M.Stefan: Thanks. I guess that is different from "Romanian Mathematical Magazine", which is the first google search result for "Romanian Mathematical Gazette". – Aryabhata Aug 01 '17 at 15:48

1 Answers1

3

Let us consider the two sequences

$$a_n = \frac{x_n}{y_n}$$

and

$$b_n = x_n y_n$$ instead

Dividing the two given recurrences (of $x_{n+1}$ and $y_{n+1}$) gives us

$$a_{n+1} = 2 b_n$$

Note that it is enough to show that $b_n$ is convergent, as that implies $a_n$ is convergent, which in turn implies $x_n^2 = a_nb_n$ is convergent, which implies $y_n$ is too.

Now multiplying the two give recurrences(of $x_{n+1}$ and $y_{n+1}$) in the question gives us

$$9 b_{n+1} = 2 + \frac{a_n}{2} + \frac{2}{a_n} = 2 + b_{n-1} + \frac{1}{b_{n-1}}$$

To simplify the algebra, Let $$s_n = \sqrt{b_n}$$ Thus $$(3s_{n+1})^2 = \left(s_{n-1} + \frac{1}{s_n}\right)^2$$

Since $s_n \gt 0$, we have

$$3s_{n+1} = s_{n-1} + \frac{1}{s_{n-1}}$$

Now by considering the even $n$ and odd $n$ separately, we just need to analyze the sequence $$3c_{n+1} = c_n + \frac{1}{c_n}$$ (for eg by setting $c_k = s_{2k}$ or $c_k = s_{2k+1}$)

Let $$L = \frac{1}{\sqrt{2}}$$

Let $$e_n = c_n -L$$

Note that $$e_n = 0 \implies e_{n+1} = 0 \implies c_n \to L$$

So assume $e_n \ne 0$

We have that $$3L = L + \frac{1}{L}$$ and so

$$3c_{n+1} -3L = c_n - L + \frac{1}{c_n} - \frac{1}{L}$$

i.e.

$$ 3 e_{n+1} = e_n - \frac{e_n}{Lc_n}$$

Thus

$$\left|\frac{e_{n+1}}{e_n}\right| = \frac{|1 - \frac{1}{Lc_n}|}{3}$$

Let us now show that $$\frac{|1 - \frac{1}{Lc_n}|}{3} \le d$$ for some constant $d \lt 1$ which implies $e_n \to 0$.

Now we have that $$c_n \ge \frac{2}{3}$$ using $x + \frac{1}{x} \ge 2$ (put $x = c_{n-1}$).

$$ \implies Lc_n \ge \frac{\sqrt{2}}{3}$$

And so

$$\frac{3Lc_n}{\sqrt{2}} \ge 1$$

$$\frac{(3-\sqrt{2})Lc_n}{\sqrt{2}} \ge 1 - Lc_n$$

Thus

$$\frac{3-\sqrt{2}}{3\sqrt{2}} \ge \frac{1 - Lc_n}{3Lc_n}$$

Now if $Lc_n \gt 1$, then $$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 - \frac{1}{Lc_n}}{3} \le \frac{1}{3}$$

if $Lc_n \le 1$ then

$$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 -Lc_n}{3Lc_n} \le \frac{3-\sqrt{2}}{3\sqrt{2}}$$

Thus $e_n \to 0$ and so $c_n \to L$. Thus $s_{2n}, s_{2n+1} \to L$ and so $b_n$ is convergent.

Aryabhata
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