Euler's Heuristic Solution to the Basel Problem

The Problem

Find the exact value of: 1 + 1/2² + 1/3² + 1/4² + 1/5² + ... = ∑(1/n²)

Euler's Heuristic Approach (c. 1735)

Step 1: The Sine Function and its Roots

The function sin(x) has roots at every integer multiple of π:

sin(x) = 0 when x = 0, ±π, ±2π, ±3π, ...

The Taylor series expansion for sin(x) is:

sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ...

Step 2: The "Infinite Polynomial" Factorization

Euler treated sin(x) as an infinite polynomial. Consider sin(x)/x:

sin(x)/x = 1 - x²/3! + x⁴/5! - x⁶/7! + ...

The roots of sin(x)/x are at x = ±π, ±2π, ±3π, ...

Using polynomial factorization, Euler wrote:

sin(x)/x = (1 - x/π)(1 + x/π)(1 - x/2π)(1 + x/2π)(1 - x/3π)(1 + x/3π)...

Combining pairs of factors:

sin(x)/x = (1 - x²/π²)(1 - x²/2²π²)(1 - x²/3²π²)(1 - x²/4²π²)...

Step 3: Expanding and Comparing Coefficients

We now have two expressions for sin(x)/x:

Series: 1 - x²/3! + x⁴/5! - x⁶/7! + ...

Product: (1 - x²/π²)(1 - x²/2²π²)(1 - x²/3²π²)...

Compare coefficients of x²:

From series: coefficient = -1/6

From product: coefficient = -1/π²(1 + 1/2² + 1/3² + 1/4² + ...)

Step 4: The Final Result

Equating the coefficients:

-1/π²(1 + 1/2² + 1/3² + 1/4² + ...) = -1/6

Multiply both sides by -π²:

1 + 1/2² + 1/3² + 1/4² + ... = π²/6

The Basel problem is solved!

Note on Rigor: This was Euler's heuristic approach. The steps weren't fully rigorous by modern standards, particularly treating sin(x) as an infinite polynomial and applying finite product rules to infinite products. However, Euler later found rigorous proofs, and this brilliant heuristic argument demonstrates his extraordinary mathematical intuition.