proof for the Ramanujan's formula ? I found this formula in a textbook in which the proof to the formula was not given
Ramanujam's formula
$$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$
Its a great equation andhow do you prove this. its a bit difficult for me and tried different methods to solve this. Iam an undergraduate and I want you to elaborate the method of solving if its complex.  
 A: More general formulation can be gotten:
$F(x) = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt{\mathrm{\cdots}}}}}$
$$F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}} $$
Which can be simplified to:
$$F(x)^2 = ax+(n+a)^2 +xF(x+n) \tag1$$
Let's assume that $F(x) = mx+k $
$deg(F(x))$ cannot be more than 1 otherwise left side degree will be bigger than right side
in Equation 1
$$F(x)^2 = ax+(n+a)^2 +xF(x+n) $$
 $$( mx+k)^2=ax+(n+a)^2+x(m(x+n)+k) $$
$$ m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$$
$m^2=m$
$m=1$ or $m=0$
$2mk=mn+k+a$
If $m=1$ then 
$2k=n+a+k$
$k=n+a$
And It is also confirming constant term of $ m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$ :
$$k^2=(n+a)^2$$
$F(x) = mx+k $
$F(x) = x + n + a $
$$  x + n + a= \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}} $$
So, setting $a=0$, $n=1$,    
$$ x+1=\sqrt{1 +x\sqrt{1 +(x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)\sqrt{1 +....}}}}}  $$
Please see Infinitely nested radicals in the wiki page as reference.
A: For all $x > 0$ and $m \in \mathbb{N}$, define $\varphi_m(x)$ by:
$$(x+1)\varphi_m(x) = \begin{cases}1,&\text{ for } m = 0\\
\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{\cdots\sqrt{1+(x+m)}}}},&\text{ otherwise. }\end{cases}$$
It is clear for $m > 0$, these functions satisfy the recurrence relations:
$$\begin{align}
     &(x+1)^2\varphi_{m}(x)^2 = 1 + x(x+2)\varphi_{m-1}(x+1)\\
\iff &\varphi_{m}(x)^2 = \varphi_{m-1}(x+1) + \frac{1 - \varphi_{m-1}(x+1)}{(x+1)^2}
\end{align}
$$
Notice for any $x > 0, y \in [0,1]$, we have $y + \frac{1-y}{(x+1)^2} \in [y,1]$. This implies
$$\varphi_{m-1}(x+1) \le \varphi_{m}(x)^2 \le 1
\quad\iff\quad
\varphi_{m-1}(x+1)^{2^{-1}} \le \varphi_{m}(x) \le 1$$
Repeat apply this to $\varphi_{m-2}(x+2), \varphi_{m-3}(x+3),\ldots$, we get:
$$\begin{align}
&\varphi_{0}(x+m)^{2^{-m}} \le \varphi_{1}(x+m-1)^{2^{-(m-1)}} \le \cdots \le \varphi_{m}(x) \le 1\\
\implies & \left(\frac{1}{x+m+1}\right)^{2^{-m}} \le \varphi_m(x) \le 1\tag{*1}
\end{align}$$
Notice for fixed $x$, the limit of L.H.S of $(*1)$ goes to $1$ as $m \to \infty$. As a consequence:
$$\sqrt{1+x\sqrt{1+(x+1)\sqrt{1 + \cdots}}} = (x+1)\lim_{m\to\infty}\varphi_m(x) = x + 1$$
A: Here is this answer adapted to this question.
Define
$$
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\dots}}}}
$$
then $f(x)^2=1+xf(x+1)$. This indicates we should look at $f(x)=x+1$.
Considering $f(x)=x+1$, we are lead to show inductively that
$$
\hspace{-18pt}x+1=\small\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\dots+\sqrt{1+(x+k-2)\sqrt{1+(x+k-1)(x+k+1)}}}}}\tag{1}
$$

Define
$$
f_{k,x}(y)=\left\{\begin{array}{}
y&\text{if }k=0\\
\sqrt{1+xf_{k-1,x+1}(y)}&\text{if }k>0
\end{array}\right.
$$
unrolled, that is
$$
f_{k,x}(y)=\small \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\dots+\sqrt{1+(x+k-2)\sqrt{1+(x+k-1)y}}}}}
$$
Note that $f_{0,x}(x+1)=x+1$. Suppose that, for some $k\ge0$, $f_{k,x}(x+k+1)=x+1$. Then
$$
\begin{align}
f_{k+1,x}(x+k+2)
&=\sqrt{1+xf_{k,x+1}(x+k+2)}\\
&=\sqrt{1+x(x+2)}\\
&=x+1
\end{align}
$$
Therefore, for all $k\ge0$, we have $f_{k,x}(x+k+1)=x+1$.
Thus $(1)$ is confirmed.

Our sequence is bounded by $f_{k,x}(1)$ and $f_{k,x}(x+k+1)$. To show that our sequence converges to $x+1$, we want to show that
$$
\lim_{k\to\infty}f_{k,x}(1)=x+1\tag{2}
$$
Note that
$$
\frac{f_{0,x+k}(x+k+1)}{f_{0,x+k}(1)}=\frac{x+k+1}{1}
$$
Suppose that, for some $j\ge0$,
$$
\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}\le(x+k+1)^{1/2^j}
$$
Then
$$
\begin{align}
\frac{f_{j+1,x+k-j-1}(x+k+1)}{f_{j+1,x+k-j-1}(1)}
&=\sqrt{\frac{1+(x+k-j-1)f_{j,x+k-j}(x+k+1)}
{1+(x+k-j-1)f_{j,x+k-j}(1)}}\\
&\le\sqrt{\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}}\\
&\le(x+k+1)^{1/2^{j+1}}
\end{align}
$$
Therefore, for all $j\ge0$,
$$
\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}\le(x+k+1)^{1/2^j}
$$
in particular, for $j=k$,
$$
\frac{f_{k,x}(x+k+1)}{f_{k,x}(1)}\le(x+k+1)^{1/2^k}
$$

Since $f_{k,x}(x+k+1)=x+1$, we have
$$
(x+1)(x+k+1)^{-1/2^k}\le f_{k,x}(1)\le (x+1)
$$
and by the squeeze theorem, we have
$$
\lim_{k\to\infty}f_{k,x}(1)=x+1
$$
confirming $(2)$, as desired.
A: $x>-1\iff \underline{x+1}=\sqrt{(x+1)^2}=\sqrt{1+2x+x^2}=\sqrt{1+x\cdot(\underline{\underline{x+2}})}$
$\begin{align}x>-2\iff \underline{\underline{x+2}}=\sqrt{(x+2)^2}=\sqrt{[(x+1)+1]^2}&=\sqrt{1+2(x+1)+(x+1)^2}=\\&=\sqrt{1+(x+1)(\underline{\underline{\underline{x+3}}})}\end{align}$
$\to x+1=\sqrt{1+x\sqrt{1+(x+1)(\underline{\underline{x+3}})}}\quad-\quad$ Can you see where this is going ? :-)
