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We define 3 sequences $(a_n),(b_n),(c_n)$ with positive terms so that $$ a_{n+1}\leq\frac{b_n+c_n}{3}\ ,\ b_{n+1}\leq\dfrac{a_n+c_n}{3}\ ,\ c_{n+1}\leq\dfrac{a_n+b_n}{3} $$ Check if any of $(a_n),(b_n),(c_n)$ converge, and if they do find their limit.

PROOF

My part of the proof is this: By adding the above inequalities we get $$ a_{n+1}+b_{n+1}+c_{n+1}\leq\frac{2}{3}(a_n+b_n+c_n) $$ We define the sequence $ (x_n) $ with $ x_n=a_n+b_n+c_n $ so we have $$ x_{n+1}\leq\frac{2}{3}x_n\Rightarrow \frac{x_{n+1}}{x_n}\leq\frac{2}{3}<1 $$ which implies that $ (x_n) $ is decreasing. We also have $ x_n>0, \forall n $ thus it's bounded, and so it converges to $ 0 $. Is it correct to say that since $ a_n<x_n $ then $ a_n\to0 $?

EDIT

I forgot to mention that we also prove that $x_n\to0$.

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    $\begingroup$ $1+1/n$ is decreasing and bounded, but do not converges to $0$ $\endgroup$ – enzotib Jun 17 '17 at 12:04
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Yes because you have the "sandwiched" inequality $0 < a_{n} < x_{n} \rightarrow 0$, so $a_{n} \rightarrow 0$.

EDIT: Of course, you have to make sure that $x_n \rightarrow 0$ in the first place, but you have that it is bounded by a geometric sequence which converges to zero which implies the stated convergence.

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  • $\begingroup$ You mean $\leq $. $\endgroup$ – Displayname Jun 17 '17 at 12:06
  • $\begingroup$ Not necessarily since $a_n$, $b_n$ and $c_n$ are assumed positive. But $\leq$ would of course also be enough, however, this way I think the difference between "converging to zero" and "being zero" becomes a bit clearer. $\endgroup$ – Andre Jun 17 '17 at 13:34
  • $\begingroup$ No, it is not equivalent to this. Take for example the case $0 < \frac{1}{n} < \frac{2}{n}$. All the inequalities are strict for every $n$, but both converge to 0 as $n \rightarrow \infty$. $\endgroup$ – Andre Jun 18 '17 at 0:45
  • $\begingroup$ That is not what you wrote. You are taking the limit since you used a $\rightarrow $. $\endgroup$ – Displayname Jun 18 '17 at 5:49
  • $\begingroup$ Your current statement literally is $ 0 < a_n < 0$. $\endgroup$ – Displayname Jun 18 '17 at 6:02
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To prove that $x_n$ converges to $0$ you need to take the limit $$ 0\leq x_{n+1}\leq \frac{2}{3}x_n \implies 0\leq l\leq \frac{2}{3}l \implies l=0 $$

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Since $\{x_n\}$ is decreasing and bounded from below, it is convergent. But this does not guarantee that the limit is zero. Indeed, since

$$0< x_{n+1}\le\frac{2}{3}x_n$$

we have

$$0< x_n\le\left(\frac{2}{3}\right)^{n-1}x_1$$

This implies that $x_n\to0$ as $n\to\infty$.

We have $\displaystyle 0<a_n<\frac{1}{3}x_n$, $\displaystyle 0<b_n<\frac{1}{3}x_n$ and $\displaystyle 0<c_n<\frac{1}{3}x_n$.

So $a_n\to 0$, $b_n\to 0$ and $c_n\to 0$ as $n\to\infty$.

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