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Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$

My try:

$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$

$\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$

$\Rightarrow x^3-2x+1=0$

and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$

So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.

how can i say that the limit is $\dfrac {-1 + \sqrt {5}}{2}$?

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We will prove that all $a_n$ are smaller than ${2 \over 3}=0.6666...$.

By induction, suppose that $0, 1/2, ... a_{n-1}, a_n < 2/3$.

then $a_{n+1} < {1 + 2/3 + 8/27 \over 3 }= {53 \over 81} < {54 \over 81} = {2\over 3}$

since $a_1 =0<{2 \over 3}$, for all n , $0 \leqslant a_n < {2 \over 3}$.

To prove the convergence we will show that a_n is increasing. Again, by induction,

$a_{n+1} - a_n = {a_n - a_{n-1} \over 3} + { (a_{n-1} - a_{n-2}) ( a_{n-1}^2 + a_{n-1} a_{n-2} + a_{n-2}^2 ) \over 3} > 0$ where

$a_4-a_3 = {1 \over 24} > 0 $ and $a_5- a_4 = {1 \over 72} > 0 $ are the two first consecutive positive differences.

We have here a strictly increasing bounded sequence i.e. a convergent one.

So the limit calculus in the question is valid. The limit is $\dfrac {-1 + \sqrt {5}}{2} \approx 0.618$

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    $\begingroup$ It seems a bit strange to me to suppose $1 < 2/3$... $\endgroup$ – CiaPan Jun 15 '18 at 13:27
  • $\begingroup$ I apologize, it was a typo, $a_1=0$, thanks $\endgroup$ – Boyku Jun 15 '18 at 14:10
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    $\begingroup$ @nbegginer how did u get this step. I tried it but can't get it. Please give hint $a_{n+1} - a_n = (a_n - a_{n-1}) { 1 + a_n^2 + a_na_{n-1} + a_2^2 \over 3} > 0$ $\endgroup$ – Girish Kumar Chandora Jun 16 '18 at 5:47
  • $\begingroup$ Oh dear me ! I review it. $\endgroup$ – Boyku Jun 16 '18 at 12:04
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Because $f(x)={1\over 3}(1+x+x^3)$ is a strictly increasing function, compute its derivative, show recursively that that $0<a_n$, remark that if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$.

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  • $\begingroup$ $a_n$ is not expression/ function of $n$. Its a recursion form , so how can you directly say by expressing into function of x. $\endgroup$ – Girish Kumar Chandora Jun 15 '18 at 8:42
  • $\begingroup$ can you explain these lines in more details. I didnt get this if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$. $\endgroup$ – Girish Kumar Chandora Jun 15 '18 at 10:29

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