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Let $\{a_n\}$ be a sequence such that $a_1=4$ and $a_{n+1}=\dfrac{5 a_n -6}{a_n -2},\, \forall n\geq 2$. Show that it converges and find its limit.

The only thing that I managed to show is that if it is convergent, the limit is either 1 or 6. I used Mathematica to see the behavior of the sequence, and I noticed that it converges to 6 and also that it is not monotonic.

I have come across some recursive sequences like this in various posts here in math.SE, but all of them where bounded and monotonic.

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If we let $b_n=a_n-6$, we get $b_{n+1}+6=\frac{5(b_n+6)-6}{b_n+4}$ which simplifies to $b_{n+1}=-\frac{b_n}{b_n+4}$. Now letting $c_n=\frac{1}{b_n}$ we get $c_{n+1}=-1-4c_n$ and $c_1=-\frac{1}{2}$. This looks almost like a geometric sequence, so finally let $d_n=c_n+\frac{1}{5}$ and we get $d_{n+1}=-4d_n$ and $d_1=-\frac{3}{10}$. Now we can see $d_n=-\frac{3}{10}(-4)^{n-1}$, and hence $$ a_n=\frac{1}{-\frac{3}{10}(-4)^{n-1}-\frac{1}{5}}+6. $$ So $a_n \to 6$ as $n \to \infty$ (the denominator grows to infinity in absolute value).

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  • $\begingroup$ Pretty impressive technique, never seen this before! Is there a textbook with similar examples? $\endgroup$
    – Laxuist
    Sep 18 '20 at 20:08
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    $\begingroup$ @Laxuist I am not aware of any specific textbook, but you might look at similar examples using the approach0.xyz/search search engine, it searches on this site as well as Art of Problem Solving site with plenty of examples to check. $\endgroup$
    – Sil
    Sep 18 '20 at 20:27
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By induction for any $n\geq1$ we obtain: $$a_{n+1}=\frac{5a_n-10+4}{a_n-2}=5+\frac{4}{a_n-2}>4$$ Thus, $a_n\geq4$ for any $n\geq1$.

Now, $$|a_{n+1}-6|=\frac{|a_n-6|}{a_n-2}\leq\frac{1}{2}|a_n-6|\leq\left(\frac{1}{2}\right)^2|a_{n-1}-6|\leq...\leq\left(\frac{1}{2}\right)^n|a_1-6|\rightarrow0.$$

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  • $\begingroup$ This was the solution I expected, but it was the hardest for me to understand. How can I get an idea of the last part: $|a_n -6|\leq \frac{1}{2} |a_{n-1} -6|$? I mean what you did, but I have no idea of how to make this thought my own. $\endgroup$
    – Laxuist
    Sep 18 '20 at 20:06
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    $\begingroup$ @Laxuist It's one of things which we need to make firstly: $a_{n+1}-6=\frac{5a_n-6}{a_n-2}-6=\frac{6-a_n}{a_n-2},$ which gives $|a_n-6|\leq\frac{1}{2}|a_{n-1}-6|.$ $\endgroup$ Sep 18 '20 at 22:30

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