Ramanujan's radical and how we define an infinite nested radical I know it is true that we have
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
The argument is to break the nested radical into something  like
$$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
However, I am not convinced. I can do something like
$$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
Something must be wrong and the reason behind should be a misunderstanding of how we define infinite nested radical in the form of 
$$ \sqrt{a_{0}+a_{1}\sqrt{a_{2}+a_{3}\sqrt{a_{4}+a_{5}\sqrt{a_{6}+\cdots}}}} $$
I researched for a while but all I could find was computation tricks but not a strict definition. Really need help here. Thanks.
 A: As others have said, the rigorous definition of an infinite expression comes from the limit of a sequence of finite terms. The terms need to be well-defined, but in practice, we just try to make sure the pattern is clear from context.
Now let's see what goes wrong with your other example. You wrote:
$$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
The problem is that each term in the sequence (e.g. if we stop at $4$) fails to include an "extra" amount (and this amount is not going to zero). So if we look at the partial sums, we see they won't converge to $4$ unless we include the extra amounts we keep pushing to the right. It's the same logical mistake as taking
\begin{align*}
  2 &= 1 + 1  \\
    &= \frac{1}{2} + \frac{1}{2} + 1  \\
    &= \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + 1 \\
    &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + 1
\end{align*}
And then saying, wait, the partial sums in the last line only converge to $1$ instead of $2$. But this is a mistake because we can't push the extra $1$ "infinitely far" right. Otherwise, the partial sum terms we write down will just look like $\frac{1}{2} + \frac{1}{4} + \cdots$ and will never include the $1$. The same thing is happening (roughly) in your example.
A: Your example doesn't quite capture the spirit of the Ramanujan expression: it arises from the succession of the integers, not from ad hoc computation to try to fit the expression to a value.
Let's make a sketch of an induction proof.  Take the base case $n=3$:
$$\sqrt{n^2}=3$$
and, for the inductive step, take the radical at the "deepest" level of the chain and apply
$$\sqrt{k^2}=\sqrt{1+(k-1)(k+1)}$$
$$=\sqrt{1+(k-1)\sqrt{(k+1)^2}}$$
this gives us exactly the sequence we need. This means we can take any $N>=3$ and construct the expression with $\sqrt{N^2}$ in the deepest radical, and know that the equality with 3 has been preserved. (What does this tell us about the limit at infinity?)
For the 4 example, there's no analogous step to get from one term of the sequence to the next; you have to compute the term at the lowest level by working through the whole chain each time.
A: $4=\sqrt{16}=\sqrt{1+3\sqrt{25}}=\sqrt{1+3\sqrt{1+4\sqrt{36}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{49}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{81}}}}}}$
$=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+\cdots}}}}}}}$
$5=\sqrt{25}=\sqrt{1+4\sqrt{36}}=\sqrt{1+4\sqrt{1+5\sqrt{49}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{81}}}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{100}}}}}}$
$=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+9\sqrt{1+\cdots}}}}}}}$
$\vdots$
$n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+(n+4)\sqrt{1+\cdots}}}}}}}$
A: The user @Eevee Trainer provided a nice explanation on how we define infinite nested radical in terms of limit of finite nested radical which should be insensitive of the starting point. For full generality in this regard, we can consider the convergence of the following finite nested radical
$$ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ 1 + \cdots n \sqrt{1 + (n+1) a_n}}}} \tag{*} $$
for a given sequence $(a_n)$ of non-negative real numbers. In this answer, I will prove the convergence of $\text{(*)}$ to $3$ under a mild condition. My solution will involve some preliminary knowledge on analysis.

Setting. We consider the map $\Phi$, defined on the space of all functions from $[0,\infty)$ to $[0, \infty)$, which is given by
$$ \Phi [f](x) = \sqrt{1 + xf(x+1)}. $$
Let us check how $\Phi$ is related to our problem. The idea is to apply the trick of computing Ramanujan's infinite nested radical not to a single number, but rather to a function. Here, we choose $F(x) = x+1$. Since
$$ F(x) = 1+x = \sqrt{1+x(x+2)} = \sqrt{1 + xF(x+1)} = \Phi[F](x), $$
it follows that we can iterated $\Phi$ several times to obtain
$$ F(x) = \Phi^{\circ n}[F](x) = \sqrt{1 + x\sqrt{1 + (x+1)\sqrt{1 + \cdots (x+n-1)\sqrt{(x+n+1)^2}}}}, $$
where $\Phi^{\circ n} = \Phi \circ \cdots \circ \Phi$ is the $n$-fold function composition of $\Phi$. Of course, the original radical corresponds to the case $x = 2$. This already proves that $\Phi^{\circ n}[F](x)$ converges to $F(x)$ as $n\to\infty$.
On the other hand, infinite nested radicals do not have any designated value to start with, and so, the above computation is far from satisfactory when it comes to defining infinite nested radical. Thus, a form of robustness of the convergence is required. In this regard, @Eevee Trainer investigated the convergence of
$$\Phi^{\circ n}[1](2) = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \cdots (n+1)\sqrt{1}}}}, $$
and confirmed numerically that this still tends to $F(2) = 3$ as $n$ grows. Of course, it will be ideal if we can verify the same conclusion for other choices of starting points and using rigorous argument.
Proof. One nice thing about $\Phi$ is that it enjoys monotonicity: For any non-negative functions $f$ and $g$ on $[0, \infty)$, the implication
$$ f \leq g \quad\implies\quad \Phi [f] \leq \Phi [g] $$
holds. From this, it is easy to establish the next result.

Lemma 1. For any $f : [0, \infty) \to [0, \infty)$, we have
$$ \liminf_{n\to\infty} \Phi^{\circ n}[f](x) \geq x+1. $$

Proof. By inductively applying the monotonicity of $\Phi$, we get
$$\Phi^{\circ (n+1)} [0](x) \geq x^{1-2^{-n}} \qquad \text{and} \qquad \Phi^{\circ n}[x](x) \geq x + 1 - 2^{-n}. $$
So, for any integer $m \geq 0$,
\begin{align*}
\liminf_{n\to\infty} \Phi^{\circ n}[f](x)
&\geq \liminf_{n\to\infty} \Phi^{\circ m}[\Phi^{\circ (n+1)}[0]](x) \\
&\geq \liminf_{n\to\infty} \Phi^{\circ m}[x^{1-2^{-(n-1)}}](x) \\
&= \Phi^{\circ m}[x](x) \\
&\geq x + 1 - 2^{-m},
\end{align*}
and letting $m \to \infty$ proves Lemma 1 as required.

Lemma 2. Suppose $f : [0, \infty) \to [0, \infty)$ satisfies
$$ \limsup_{x\to\infty} \frac{\log \log \max\{e, f(x)\}}{x} < \log 2. $$
Then it follows that
$$ \limsup_{n\to\infty} \Phi^{\circ n}[f](x) \leq x+1. $$

Proof. By the assumption, there exists $C > 1$ and $\alpha \in [0, 2)$ such that $f(x) \leq C e^{\alpha^x} (x + 1)$. By appealing to the monotonicity of $\Phi$ again, we will show that
$$ \Phi^{\circ n}[f](x) \leq C^{2^{-n}} e^{(\alpha/2)^n \alpha^x} (x+1) $$
holds for all $n \geq 0$. To this end, we invoke the principle of mathematical induction. This is certainly true in the base case $n = 0$. Furthermore, assuming that this is true for $n \geq 0$,
\begin{align*}
\Phi^{\circ (n+1)}[f](x)
&\leq \Phi\left[C^{2^{-n}} e^{(\alpha/2)^n \alpha^x} (x+1)\right](x) \\
&= \left( 1 + x C^{2^{-n}} e^{(\alpha/2)^n \alpha^{x+1}} (x+2) \right)^{1/2} \\
&\leq C^{2^{-n-1}} e^{(\alpha/2)^{n+1} \alpha^{x}} (x+1),
\end{align*}
establishing the induction step and hence the claim. Therefore, letting $n \to \infty$ proves the desired result.

Corollary. Suppose $a_n \geq 0$ satisfies
$$ \limsup_{n\to\infty} \frac{\log\log \max\{e, a_n\}}{n} < \log 2. $$
Then for any $x \geq 0$, we have
$$ \lim_{n\to\infty} \sqrt{1 + x \sqrt{1 + (x+1) \sqrt{ 1 + \cdots + (x+n-2) \sqrt{1 + (x+n-1) a_n }}}} = x+1. $$

Proof. Apply Lemma 1 and 2 to the function $f$ that interpolates the sequence $(a_n)$, such as using piecewise linear interpolation.

Remark. The bound in Lemma 2 and Corollary is optimal. Indeed, consider the non-example in OP's question of expanding $4$ as in Ramanujan's infinite nested radical. So we define the sequence $a_n$ so as to satisfy
$$ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ 1 + \cdots n \sqrt{1 + (n+1) a_n}}}} = 4. $$
Its first 4 terms are given as follows.
$$ a_1 = \frac{15}{2}, \quad 
a_2 = \frac{221}{12}, \quad
a_3 = \frac{48697}{576}, \quad
a_4 = \frac{2371066033}{1658880}, \quad \cdots. $$
Then we can show that $\frac{1}{n}\log \log a_n \to \log 2$ as $n \to \infty$.
