On Ramanujan’s Nested Radicals

In 1911, a 23-year-old clerk from Madras posed a problem in the Journal of the Indian Mathematical Society that stumped professional mathematicians for months. The answer? A simple “3.” But getting there required a formula so elegant it’s still studied today. Welcome to the world of Ramanujan’s nested radicals.

What Are Nested Radicals?

A nested radical is a square root inside a square root inside a square root—Russian dolls made of math. They look intimidating, but Ramanujan (1887–1920) discovered formulas that tame them completely.

Here’s the thing: nested radicals aren’t just mathematical curiosities. They have real applications in Number Theory and Numerical Methods, particularly for computing limits and approximating irrational numbers.

Let me walk you through Ramanujan’s derivation. It starts with something you already know.

The Derivation: From Binomial Theorem to Infinite Radicals

Step 1: Start with the Binomial Square

The binomial theorem for degree 2:

$$ (x + a)^2 = x^2 + 2xa + a^2 \tag{1} $$

Nothing fancy yet. Now here’s Ramanujan’s clever move.

Step 2: Make a Strategic Substitution

Replace \( a \) with \( (n + a) \), where \( x, n, a \in \mathbb{R} \):

$$ (x + n + a)^2 = x^2 + 2x(n + a) + (n + a)^2 $$

Expand and rearrange:

$$ (x + n + a)^2 = x^2 + 2xn + 2ax + (n + a)^2 $$

Step 3: Rearrange Strategically

Group terms in a specific way (this is where the magic happens):

$$ (x + n + a)^2 = ax + (n + a)^2 + x^2 + 2xn + ax $$

$$ = ax + (n + a)^2 + x(x + 2n + a) $$

Take the square root of both sides:

$$ x + n + a = \sqrt{ax + (n + a)^2 + x(x + 2n + a)} \tag{2} $$

Key insight: Notice that \( (x + 2n + a) \) appears inside the radical. We can apply the same formula to THIS expression. That’s how the nesting begins.

Step 4: Apply the Formula Recursively

Observe that \( x + 2n + a = (x + n) + n + a \).

So if we replace \( x \) with \( x + n \) in equation (2):

$$ x + 2n + a = \sqrt{a(x + n) + (n + a)^2 + (x + n)(x + 3n + a)} \tag{3} $$

Keep going:

$$ x + 3n + a = \sqrt{a(x + 2n) + (n + a)^2 + (x + 2n)(x + 4n + a)} \tag{4} $$

$$ x + 4n + a = \sqrt{a(x + 3n) + (n + a)^2 + (x + 3n)(x + 5n + a)} \tag{5} $$

The general pattern for any \( k \in \mathbb{N} \):

$$ x + kn + a = \sqrt{ a(x + (k-1)n) + (n + a)^2 + (x + (k-1)n)(x + (k+1)n + a) } \tag{6} $$

Step 5: Substitute and Nest

Now substitute equation (3) into equation (2):

$$ x + n + a = \sqrt{ ax + (n + a)^2 + x \sqrt{ a(x + n) + (n + a)^2 + (x + n)(x + 3n + a) }} \tag{7} $$

Substitute equation (4) into equation (7):

$$ x + n + a = \sqrt{ ax + (n+a)^2 + x\sqrt{ a(x+n) + (n+a)^2 + (x+n)\sqrt{ a(x+2n) + (n+a)^2 + (x+2n)(x+4n+a) }}} \tag{8} $$

The General Formula: Ramanujan’s Nested Radical

Continuing this process gives us the general formula for \( k \)-nested radicals:

$$ x + n + a = \sqrt{ ax + (n+a)^2 + x\sqrt{ a(x+n) + (n+a)^2 + (x+n)\sqrt{ \cdots }}} $$

More precisely, with \( k \) levels of nesting:

Special Cases: Where the Magic Happens

The general formula is powerful, but the special cases are where you get the famous results.

Case 1: The Infinite Self-Similar Radical (\( n = 0 \))

Setting \( n = 0 \) in the general formula:

$$ x + a = \sqrt{ ax + a^2 + x\sqrt{ ax + a^2 + x\sqrt{ ax + a^2 + x\sqrt{ \cdots }}}} \tag{11} $$

The pattern repeats infinitely! This is a self-similar nested radical.

Case 2: Ramanujan’s Famous Problem (\( x = 1, a = 0 \))

This is the one that stumped mathematicians in 1911. Setting \( x = 1 \) and \( a = 0 \):

$$ 1 + n = \sqrt{ n^2 + \sqrt{ n^2 + (1+n)\sqrt{ n^2 + (1+2n)\sqrt{ \cdots }}}} \tag{12} $$

From equation (8) with these values:

$$ 1 + n = \sqrt{ n^2 + \sqrt{ n^2 + (1+n)\sqrt{ n^2 + (1+2n)(1+4n) }}} \tag{13} $$

Case 3: The Famous “3” Problem (\( x = a = n \))

Setting \( x = a = n \):

$$ 3n = \sqrt{ 5n^2 + n\sqrt{ 6n^2 + 2n\sqrt{ 7n^2 + 3n\sqrt{ \cdots }}}} \tag{14} $$

Now set \( n = 1 \):

This is the problem Ramanujan posed in 1911. The answer is simply 3, but proving it requires the machinery we just developed.

Case 4: Formula for Even Numbers (\( x = n, a = 0 \))

Setting \( x = n \in \mathbb{N} \) and \( a = 0 \):

$$ 2n = \sqrt{ n^2 + n\sqrt{ n^2 + 2n\sqrt{ n^2 + 3n\sqrt{ \cdots }}}} \tag{16} $$

Case 5: Formula for Odd Numbers (\( x = n, a = 1 \))

Setting \( x = n \in \mathbb{N} \) and \( a = 1 \):

$$ 2n + 1 = \sqrt{ n + (n+1)^2 + n\sqrt{ 2n + (n+1)^2 + 2n\sqrt{ 3n + (n+1)^2 + \cdots }}} $$

Simplified:

$$ 2n + 1 = \sqrt{ n + (n+1)^2 + n\sqrt{ 2n + (n+1)^2 + 2n\sqrt{ \cdots + \sqrt{ (k+3)n^2 + (k+3)n + 1 }}}} \tag{17} $$

The Symmetry Property

Here’s something beautiful: since \( x, n, a \) are all real numbers, you can swap them around.

The formula \( x + n + a \) can be computed by:

• Starting with \( ax \) and incrementing by \( n \)
• Starting with \( an \) and incrementing by \( x \)
• Starting with \( xa \) and incrementing by \( n \)

All three approaches yield the same result—just with different intermediate expressions inside the radicals. This symmetry is a hallmark of Ramanujan’s work: deep simplicity hiding beneath apparent complexity.

Quick Reference: Special Cases Summary

Frequently Asked Questions

The Bottom Line

Ramanujan took the humble binomial theorem and, through one clever substitution and recursive thinking, unlocked a general formula for infinite nested radicals. The result: any expression \( x + n + a \) can be written as a nested radical, and famous “impossible” problems like \( \sqrt{5 + \sqrt{6 + 2\sqrt{7 + \cdots}}} = 3 \) become trivial corollaries.

That’s the Ramanujan way: find the pattern, generalize ruthlessly, and watch specific puzzles dissolve into special cases.