Proving convergence of sequence with induction I have a sequence defined as $a_{1}=\sqrt{a}$ and $a_{n}=\sqrt{1+a_{n-1}}$ and I need to prove that it has an upper bound and therefore is convergent. So i have assumed that the sequence has a limit and by squaring I got that the limit is $\frac{1+\sqrt{5}}{2}$ only $ \mathbf{if}$ it converges. 
What methods are there for proving convergence? I am trying to show that $a_{n}<\frac{1+\sqrt{5}}{2}$ by induction but could use some help since I have never done induction proof before.
Progress:
Step 1(Basis): Check if it holds for lowest possible integer: Since $a_{0}$ is not defined, lowest possible value is $2$.
$a_{2}=\sqrt{1+a_{1}}=\sqrt{1+\sqrt{a}}=\sqrt{1+\sqrt{\frac{1+\sqrt{5}}{2}}}< \frac{1+\sqrt{5}}{2}$. 
Step 2: Assume it holds for $k\in \mathbb{N},k\geq 3$. If we can prove that it holds for $n=k+1$ we are done and therefore it holds for all  $k$.
This is were i am stuck:  $a_{k+1}=\sqrt{1+a_{k}}$. I don't know how to proceed because I don't know where I am supposed to end. 

Let $f(x) = \sqrt{1+x}-x$. We find that $f'(x) = \frac 1 {2\sqrt{1+x}} -1 <0$. This means that $f$ is a strictly decreasing function. Set $\phi = \frac{1+\sqrt 5}{2}$.
We now that $f(\phi)=0$. We must then have that $f(x)>0$ if $x<\phi$ and $f(x)<0$ if $x>\phi$. So $a_{n+1}>a_n$ if $a_n< \phi$ and $a_{n+1} < a_n$ if $a_n > \phi$.  
I claim that if $a_1<\phi$ then $a_n < \phi$ for all $n$. This is proven by induction. Assume that $a_k < \phi$. Then $a_{k+1}^2 = a_k +1 < 1+\frac{1+\sqrt 5}{2}= \frac{6+2\sqrt{5}}{4} =\phi^2$. So $a_{k+1} < \phi$ and by induction we get that $a_n < \phi$ for all $n$. 
If $a_1<\phi$ we thus know that we get a bounded increasing sequence. All bounded increasing sequences converge. To deal with the case when $\sqrt {a_1}>\phi$ is left as an exercise. 
