Limit involving fibonacci series with an integral If $$I_1=\int _0 ^1 \frac {dx}{1+\sqrt {x }}, \hspace{5mm} I_2=\int _0 ^1 \frac {dx}{1+\frac {1}{1+\sqrt {x}}}, \hspace{5mm} I_3=\int _0 ^1 \frac {dx}{1+\frac {1}{1+\frac {1}{1+\sqrt {x}}}} $$ then $$ \lim _{n \to \infty} I_n \hspace{5mm}\text{ is equal to ?} $$ 
I simplified all these integrals and found the numbers appearing in numerator and denominator are in fibonacci series. Thus the general form is 
$$I_n=\int_{0}^{1} \left(\frac{F_{n+1} + F_{n} \, \sqrt{x}}{F_{n+2} + F_{n+1} \, \sqrt {x}} \right) \, dx$$  where $$F_{n} \text { is nth fibonacci number}$$ 
But I don't know how to complete this. Thanks!
Note of editing: Originally $(n+1)_{f}$ was given as the $n^{th}$ Fibonacci number. To make use of standard notation the notation was changed to $F_{n}$. 
 A: Since $\sqrt{x}$ is increasing, positive and bounded by $1$ on the interval $(0,1)$, $I_n$ is bounded between 
$$ [0;\underbrace{1,1,\ldots,1}_{n\text{ times}}]\quad\text{and}\quad [0; \underbrace{1,1,\ldots,1}_{n-1\text{ times}},2]$$
so by squeezing $I_n$ converges to $[0;1,1,1,1,\ldots]=\frac{-1+\sqrt{5}}{2}$, since two consecutive convergents $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$ of the same ordinary continued fraction are separated by a distance equal to $\frac{1}{q_n q_{n+1}}$, and the sequence $\{q_n\}_{n\geq 1}$ has an exponential growth.
Here I am using the compact notation for ordinary continued fractions: $[a;b,c,d]$ means $a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}.$
A: It is well known that$$\lim_{n \rightarrow \infty}\frac{(n+1)_f}{n_f}=\phi$$where $\phi$ is golden ratio, $\frac{\sqrt{5}+1}{2}$. Therefore,$$\lim_{n \rightarrow \infty}I_n=\lim_{n \rightarrow \infty}\frac{1}{\phi}=\frac{\sqrt{5}-1}{2}$$
A: Divide out $(n+1)_f$ from denominator and numerator to make it look better, as
$$I_n=\int_0^1 \frac{1+b_n\sqrt{x}}{a_n+\sqrt{x}}dx$$
where $a_n = (n+2)_f/(n+1)_f , b_n = (n)_f/(n+1)_f$.
Then we integrate using substitution $y^2=x$, which turns it into an integral of a rational function, which is solvable with elementary functions. I'm lazy so I just fed it to Mathematica to get: 
$$I_n=\int_0^1 \frac{1+b_n\sqrt{x}}{a_n+\sqrt{x}}dx = -2 (-1 + a_n b_n) \sqrt{x} + b_n x + 2 a_n (-1 + a_n b_n) \log{(a_n + \sqrt{x})}\mid^1_0$$
$$= -2 a_n b_n+2 a_n (a_n b_n-1) (\log (a_n+1)-\log (a_n))+b_n+2$$
We know that the ratio between successive terms of Fibonacci series approaches the golden ratio $$\lim_{n\rightarrow \infty} a_n = \phi = \frac{1+\sqrt{5}}{2}, \lim_{n\rightarrow \infty} b_n = 1/\phi$$ So we take the limit to $a=\phi, b=1/\phi$ to get $$\lim_{n\rightarrow \infty}I_n=2/(1 + \sqrt{5}) = 1/\phi$$
A: It's well known that
$$1+\frac {1}{1+\frac {1}{1+\frac {1}{1+\cdots}}}\to\varphi$$
Therefore,
$$\lim _{n \to \infty} I_n = \frac{1}{\varphi}$$
