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Let ($a_n$) be a sequence with $a_n \in(\frac{1}{2},1)$ for all $n\geq0$ Define the sequence $(x_n)_{n=1}^{\infty}$ by $x_0=a_0$ and $x_{n+1}=\frac{a_{n+1}+x_n}{1+a_{n+1}x_n}$. Does the sequence diverge? If not what are the possible value of $lim_{x\rightarrow \infty}x_n$?

I can connect the sequence with the hyperbolic tan function but I failed to proceed further

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$$x_n=\tanh(y_n) ;\ a_n=\tanh(b_n)$$

Then we have $$y_{n+1}=y_n+b_{n+1} $$

Since $b_{n+1}>\tanh^{-1}(1/2)>0$, then $y_{n+1}>y_n+\tanh^{-1}(1/2)$

So $y_n\rightarrow\infty$. Therefore $x_n\rightarrow1$

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  • $\begingroup$ I'm dealing with the test of the International Mathematics Competition for University Students, 2011, and I've had a lot of difficulties, so I hope someone could help me to discuss the questions. I've posted the question 2 (first day) before, in math.stackexchange.com/questions/2924555/… and this one is the question 1 (second day). Sorry, I couldn't urdestand your answer. What are $y_n,b_n$? $\endgroup$ – Na'omi Sep 21 '18 at 15:29
  • $\begingroup$ The only thing I could get before: Note that, for $n\geq1$, $\dfrac{1/2+x_n}{1+x_n}<x_{n+1}<\dfrac{1+x_n}{1+x_n/2}$ So, if the sequence converges to $L$, we have $\dfrac{1/2+L}{1+L}<L<\dfrac{1+L}{1+L/2}$ It means i) $\dfrac{1}{2}+L<L+L^2$ $L^2>\dfrac{1}{2}$ $L>\dfrac{1}{\sqrt2}$ ii) $L<\dfrac{1+L}{1+L/2}$ $L+L^2/2<1+L$ $L^2<2$ $L<\sqrt2$ So, if the sequence converges, its limite is in $\bigg(\dfrac{1}{\sqrt2},\sqrt2\bigg)$. $\endgroup$ – Na'omi Sep 21 '18 at 15:31
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    $\begingroup$ @Na'omi $y_n$ and $b_n$ are new sequences I defined to make the problem easier. I defined them implicitly but an explicit definition $y_n = \tanh^{-1} (x_n)$, $b_n = \tanh^{-1} (a_n)$ works just as well. $\endgroup$ – Kitegi Sep 22 '18 at 16:54
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    $\begingroup$ @Na'omi I chose it because I know that $\tanh(a+b) = \frac{\tanh(a) +\tanh(b)}{1+\tanh(a)\tanh(b)}$. So it was the first thing I tried. Here's a simpler example to get a better idea. Suppose we had $x_{n+1} = x_n a_n$. If we set $x_n,a_n = \exp(y_n),\exp(b_n)$ we get $y_{n+1} = y_n + b_n$. It's easier to think of using the $\exp$ function here than it is to think of $\tanh$ in the original question, but the idea is the same. Even if I didn't happen to think of this trick, the answer would still be manageable, just a little longer. (cont. in next comment) $\endgroup$ – Kitegi Sep 24 '18 at 20:22
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    $\begingroup$ @Na'omi You could take your idea of taking the limit, but instead of using the two bounds of $a_{n+1}$ to bound the denominator from above and the nominator from below (or the other way around), you could try to get rid of the $a_{n+1}$ in the denominator first, like this: $$\frac{a+x}{1+ax} = \frac{a+x}{x(a+1/x)} = \frac{1}{x} \left(1+\frac{x-1/x}{a+1/x}\right) $$ This should give you a tighter bound, and hopefully lead to the answer. $\endgroup$ – Kitegi Sep 24 '18 at 20:27

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