Calculation of Limit of a Repeating Continued Fraction I'm sorry if this is a duplicate. I have no idea on what kind of "name" i should give to this, and therefore i have no idea on how to search on the internet for help on understanding it. If it happens that this is a duplicate, i would be grateful if you could link me to where there are any solutions for this.
I need to prove for an exercise on my analysis book that the following sequence
$$
{\cfrac{1}{1+\cfrac{1}{5}}},\quad
{\cfrac{1}{1+\cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{5}}}}},\dotsc
$$
is monotone and converges to ${\frac{-5+\sqrt{45}}{2}}$
I imagine that once i get on how to determine it's limit, it will be easy to prove that it is in fact monotone. I have no idea on how to approach it though. Any tips?
 A: 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 $t^2 + 5t-5.$
=========================================
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
=========================================

A: 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}

A: 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.
