Ramanujan's radical and how we define an infinite nested radical [duplicate]

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.

Solution 1:

Introduction:

The issue is what "..." really "represents."

Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."

This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.

On Infinite Sums:

A simple example: the sum of the reciprocals of squares:

$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$$

This is a well known summation. It is the Riemann zeta function $\zeta(s)$ at $s=2$, and is known to evaluate to $\pi^2/6$ (proved by Euler and known as the Basel problem).

Another, easier-to-handle summation is the geometric sum

$$1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ...$$

This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.

Another geometric series you might see in proofs that $0.999... = 1$ is

$$9 \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1,000} + \frac{1}{10,000} + ... \right)$$

which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by

$$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$

So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.

Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.

So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum

$$\sum_{k=1}^\infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$

for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:

$$S(n) = \sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$

Then we get a sequence of sums:

$$S(1), S(2), S(3), ...$$

or equivalently

$$x_1 \;\;,\;\; x_1 + x_2\;\;,\;\; x_1 + x_2 + x_3\;\;,\;\; ...$$

Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")

For the case of our first geometric sum, we immediately see the sequence of partial sums

$$1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8},...$$

Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is

$$S(n) = \frac{2^{(n+1)}-1}{2^{n}}$$

We can then easily consider the limit of these partial sums:

$$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} \frac{2^{(n+1)}-1}{2^{n}} = \lim_{n\to\infty} \frac{2^{(n+1)}}{2^{n}} - \frac {1}{2^{n}} = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^{n}}$$

Obviously, $1/2^{n} \to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude

$$\lim_{n\to\infty} S(n) = \lim_{n\to\infty} 2 - \lim_{n\to\infty} \frac{1}{2^n} = 2 - 0 = 2$$

And thus we say

$$\sum_{k=0}^\infty \left(\frac 1 2 \right)^k = 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + ... = 2$$

because the partial sums approach $2$.

On Continued Fractions:

That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.

In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:

$$1 = \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}}$$

As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?

For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.

What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} \; 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents

$$\frac{1}{2} , \frac{1}{2-\frac{1}{2}}, \frac{1}{2-\frac{1}{2-\frac{1}{2}}},...$$

Working out the numbers, we find the sequence to be

$$\frac{1}{2},\frac{2}{3},\frac{3}{4},...$$

Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form

$$\frac{n-1}{n}$$

Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:

$$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} \frac{n-1}{n} = \lim_{n\to\infty} 1 - \frac 1 n = \lim_{n\to\infty} 1 - \lim_{n\to\infty} \frac 1 n = 1 - 0 = 1$$

Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!

On Infinite Radicals:

So now, we touch on infinite nested radicals. They're messier to deal with but doable.

One of the simpler examples of such radicals to contend with is

$$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$

As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $n\to\infty$.

Nested radicals are a lot messier than the previous, but we manage.

So first let the sequence of convergents be given by cutting off everything after the $n^{th} \; 2$ in the expression. Thus we get the sequence

$$\sqrt 2 \;\;,\;\; \sqrt{2 + \sqrt{2}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2}}}\;\;,\;\; \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)

$$C(n) = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$

(I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)

So luckily, then, we can find the limit of $C(n)$:

$$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right)$$

It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus

$$\lim_{n\to\infty} C(n) = \lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^{n+1}}\right) = 2\cos(0) = 2\cdot 1 = 2$$

Thus, since its convergents approach $2$, we can conclude that

$$2 = \sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +\sqrt{2 +...}}}}}$$

A Lengthy Conclusion:

So, in short, how do we evaluate an infinite expression, be it radical, continued fraction, sum, or otherwise?

We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $n\to\infty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.

This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.

This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)

Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution.

(How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$. This logic also relies on the assumption that there is only one solution to a given expression that is valid. Depending on the context, you might see cases where multiple solutions are valid but this "approaching by hand" method will only get you some of the solutions. This touches on the notion of "unstable" and "stable" solutions to dynamical systems - which I believe is the main context where such would pop up - but it's a bit overkill to discuss that for this post.)

So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.

We begin with the radical itself:

$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$

Let us begin by getting a series of convergents:

$$\sqrt{1} \;\;,\;\; \sqrt{1 + 2\sqrt{1}} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3\sqrt{1}}} \;\;,\;\;$$

Because the $\sqrt{1}$ isn't necessary, we just let it be $1$.

$$1 \;\;,\;\; \sqrt{1 + 2} \;\;,\;\; \sqrt{1 + 2\sqrt{1 + 3}} \;\;,\;\;$$

Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then

• $C(1) = 1$
• $C(2) \approx 1.732$
• $C(3) \approx 2.236$
• $C(4) \approx 2.560$
• $C(5) \approx 2.755$
• $C(6) \approx 2.867$
• $C(7) \approx 2.929$
• $C(8) \approx 2.963$
• $C(9) \approx 2.981$
• $C(10) \approx 2.990$

To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result

$$C(50) \approx 2.999 \; 999 \; 999 \; 999 \; 99$$

So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like

$$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$

Whew! Hopefully that lengthy post was helpful to you!

A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.

Solution 2:

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 starting points and using rigorous argument.

Proof. One nice thing about $\Phi$ is that it enjoys monotone, i.e., if $f \leq g$, then $\Phi [f] \leq \Phi [g]$. From this, it is easy to establish the following observation.

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

Proof. We inductively apply the monotonicity of $\Phi$ to find that

$$\Phi^{\circ (n+1)} [0](x) \geq x^{1-2^{-n}} \qquad \text{and} \qquad \Phi^{\circ n}[x](x) \geq x + 1 - 2^{-n}. $$

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. If $\limsup_{x\to\infty} \frac{\log \log \max\{e, f(x)\}}{x} < \log 2$, then $\limsup_{n\to\infty} \Phi^{\circ n}[f](x) \leq x+1$.

Proof. By the assumption, there exists $C > 1$ and $\alpha \in [0, 2)$ such that $f(x) \leq C e^{\alpha^x} (x + 1)$. Again, applying monotonicity of $\Phi$, we check that $\Phi^{\circ n}[f](x) \leq C^{2^{-n}} e^{(\alpha/2)^n \alpha^x} (x+1)$ holds. This is certainly true if $n = 0$. Moreover, assuming that this is true for $n$,

\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*}

Now letting $n \to \infty$ proves the desired result.

Corollary. If $a_n \geq 0$ satisfies $\limsup_{n\to\infty} \frac{\log\log \max\{e, a_n\}}{n} < \log 2$, then for any $x \geq 0$,

$$ \lim_{n\to\infty} \sqrt{1 + x \sqrt{1 + (x+1) \sqrt{ 1 + \cdots + (x+n-2) \sqrt{1 + (x+n-1) a_n }}}} = x+1. $$

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

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 $n \to \infty$.