Calculation of Limit of a Repeating Continued Fraction
Let $a_0=0$ and $a_{n+1}=\cfrac1{1+\cfrac1{5+a_n}}$.
Prove by induction:
1) $a_n<\dfrac{-5+\sqrt{45}}2$
2) $a_{n+1}>a_n$ i.e. monotone
So that it converges and that it must converge to some $a'$ such that:
$$a'=\cfrac1{1+\cfrac1{5+a'}}$$
(feel free to ask if you need more tips on any steps below, hover on the below tips to see major steps)
Induction step on proving 1)
\begin{align}&a_0<\cfrac{-5+\sqrt{45}}2\\&a_{n+1}=\cfrac1{1+\cfrac1{5+a_n}}<\cfrac1{1+\cfrac1{5+\cfrac{-5+\sqrt{45}}2}}=\cfrac{-5+\sqrt{45}}2\end{align}
Induction on 2)
\begin{align}&a_1>a_0\\&a_{n+1}=\cfrac1{1+\cfrac1{5+a_n}}>\cfrac1{1+\cfrac1{5+a_{n-1}}}=a_n\end{align}
Here is the method of Lagrange and Gauss for this: the "reduced" quadratic form $x^2 + 5x - 5 y^2$ is in a very short cycle. Notice that the limit is the positive root of $t^2 + 5t-5.$
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 5 -5
0 form 1 5 -5
1 0
0 1
To Return
1 0
0 1
0 form 1 5 -5 delta -1 ambiguous
1 form -5 5 1 delta 5 ambiguous
2 form 1 5 -5
form 1 x^2 + 5 x y -5 y^2
minimum was 1rep x = 1 y = 0 disc 45 dSqrt 6 M_Ratio 36
Automorph, written on right of Gram matrix:
-1 -5
-1 -6
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Hint : Simply consider the difference between two successive recurring terms and then it is easy to show that the sequence is monotonically increasing.
Now, $$\frac{1}{5} > 0\\\implies1+ \frac{1}{5} > 1\\\frac{1}{1 +\frac{1}{5}} < 1$$.
Use this argument several times and try to deduce that the given sequence is bounded above by 1 and then by monotone-bounded theorem, you know that there exists a limit of the sequence. Let it be l. Now a routine computation yields your answer.
N.B.: in order to show that the sequence is monotone , I guess you will need to show by induction that $$a_n < \frac{-5 + \sqrt{45}}{2}$$. And this can also come handy to show that the sequence is bounded above.