Nested Radical of Ramanujan

Question

I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.

$$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$

I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:

Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$

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and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$

Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)<x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:

For $n=1$, $a_1(x)=\sqrt{1+x}<1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n>1$.

Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:

Claim: $l(x)=x+1\quad \forall i\ge 0$

Proof: Fix $x$. Let $l(x)<x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon>0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)<\epsilon$$, and if that is true then $$a_n(x)>x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) < \epsilon$$ Now, \begin{align} x+1-a_n(x) = & x+1-\sqrt{1+xa_{n-1}(x+1)} \\ \ =& \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ =& x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ <& \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\ \ <& \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\ \ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\ \ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\ \ <& \frac{x(x+1)}{x+n-1} \end{align}

Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}<\epsilon \Rightarrow x< \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.

Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved.

Answer 1

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 $a_n(x)\to x+1$.

Answer 2

$x>-1\iff \underline{x+1}=\sqrt{(x+1)^2}=\sqrt{1+2x+x^2}=\sqrt{1+x\cdot(\underline{\underline{x+2}})}$

$\begin{align}x>-2\iff \underline{\underline{x+2}}=\sqrt{(x+2)^2}=\sqrt{[(x+1)+1]^2}&=\sqrt{1+2(x+1)+(x+1)^2}=\\&=\sqrt{1+(x+1)(\underline{\underline{\underline{x+3}}})}\end{align}$

$\to x+1=\sqrt{1+x\sqrt{1+(x+1)(\underline{\underline{x+3}})}}\quad-\quad$ Can you see where this is going ? :-)