Nested Radical of Ramanujan

Variant of your proof, for some clarity.

Define $a_1(x)=\sqrt{1+x}$ and $a_{n+1}(x)=\sqrt{1+xa_n(x+1)}$.

Define $b_n(x)=1+x-a_n(x)$. By your proof, we know that $b_n(x)$ is decreasing and bounded below by zero. You want to show that $b_n(x)\to 0$, and then you are done.

Now (this is pretty much exactly your proof, but it is made clearer by have $b_n$ defined): $$\begin{align}b_{n+1}(x) &= 1+x - a_{n+1}(x) \\&=\frac{(1+x)^2-a_{n+1}(x)^2}{1+x+a_{n+1}(x)} \\&=\frac{1+2x+x^2-(1+xa_n(x+1))}{1+x+a_{n+1}(x)} \\&= \frac{xb_n(x+1)}{1+x+a_{n+1}(x)} \\&\leq\frac{x}{2+x}b_{n}(x+1) \end{align}$$

By induction (for $n>k\geq 1$) you can show that:

$$b_n(x) \leq \frac{x(x+1)}{(x+k)(x+k+1)}b_{n-k}(x+k)$$

Therefore, for $k=n-1$, we get:

$$0\leq b_n(x)\leq \frac{x(x+1)}{(x+n-1)(x+n)}b_1(x+n-1) \leq \frac{x(x+1)}{x+n}$$

Therefore, $b_n(x)\to 0$, and hence $a_n(x)\to x+1$.