Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem

Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times  }} = 3$

After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.
So my question is

Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.

Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .
If the solution from that book is required I can post it but it is not complete as I mentioned. 
Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.
 A: $x + 1 = \sqrt {1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{}.....}}}$. Put $x=2$ gives you the solution. For proof see http://zariski.files.wordpress.com/2010/05/sr_nroots.pdf 
A: Note that $\frac{x+y}{x+z} < \frac{y}{z}$ for positive $x, y, z$; thus $\frac{\sqrt{x+y}}{\sqrt{x+z}} < \frac{\sqrt{y}}{\sqrt{z}}$ for $y > z$.
Thus we get $\frac{a_{n+1}}{a_n} = \frac{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1+(n+1)\sqrt{1}}}}}}}{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}} < \frac{\sqrt{\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1+(n+1)\sqrt{1}}}}}}}{\sqrt{\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}} < \frac{\sqrt{\sqrt{\cdots\sqrt{n+2}}}}{\sqrt{\sqrt{\cdots\sqrt{1}}}} = O(n^{(2^{-n})})$.
Next, $\ln{a_{n+1}}-\ln{a_n} = O(\frac{\ln_n}{2^n})$. Summing the equations we get $\ln{a_n} - \ln{a_1} = O(\frac{\ln_1}{2^1} + \cdots + \frac{\ln_n}{2^n})$; letting $n$ to approach $\infty$, we get $\ln{a} - \ln{a_1} = O(1)$, thus there is a finite limit of $a_i$.
A: Note that
$$
x+1=\sqrt{1+x(x+2)}\tag{1}
$$
Iterating $(1)$, we get
$$
x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\color{#C00000}{(x+5)}}}}}\tag{2}
$$
Note that 
$$
s_3=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\color{#C00000}{\sqrt{1}}}}}}\tag{3}
$$
Instead of $\color{#C00000}{\sqrt{1}}$ as in the last term of $(3)$, $(2)$ has $\color{#C00000}{(x+n+2)}$. Thus, the increasing sequence in $(3)$ is bounded above by $x+1$. Thus, the sequence in $(3)$ has a limit.
A: Let $f_n(0)=\sqrt{1+n}$ and $f_n(k)=\sqrt{1+(n-k)f_n(k-1)}$. Then $0<f_n(0)<n+1$ when $n>0$. Assume that $f_n(k)<n+1-k$ and we can show by induction that
$$
f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1)
$$
for all k. Your expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$, so converges.
A: This is a known question of Putnam competition: I give a complete proof with details:
$f_n(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+n-1)\sqrt{1+(x+n)}}}}$
it is clear that for $m>n$, we have $f_m(x)>f_n(x)$. Moreover 
$x+1=\sqrt{1+x(x+2)}=\sqrt{1+x\sqrt{1+(x+1)(x+2)}}=\cdots=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\cdots+(x+n-1)(x+n+1)}}}\geq f_n(x)$.
So this tells us that $\lim_{n\to \infty}f_n(x)$ exists. Now let $\lim_{n\to \infty}f_n(x)=f(x)$ hence $f(x)\leq x+1$ and $f(x)>\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+...}}}}=\frac{x+\sqrt{x^2+4}}{2}>x$ hence $x<f(x)<x+1$. Now, let $g(x)=x+1-f(x)$ then $0\leq g(x)<1$, it is easy to see that $(f(x))^2=1+xf(x+1)$ hence $(x+1+f(x))g(x)=xg(x+1)$, so 
$$\frac{g(x)}{x}\leq \frac{g(x+1)}{x+1}\leq \cdots\leq \frac{g(x+n)}{x+n}<\frac{1}{x+n}\to 0$$ hence $f(x)=x+1$, so $f(2)=3$
