# Does this recursive sequence converge (non monotonic)?

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.

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).

• Pretty impressive technique, never seen this before! Is there a textbook with similar examples? Sep 18 '20 at 20:08
• @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.
– Sil
Sep 18 '20 at 20:27

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

• 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. Sep 18 '20 at 20:06
• @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 $|a_n-6|\leq\frac{1}{2}|a_{n-1}-6|.$ Sep 18 '20 at 22:30