Basel Problem - Euler's Approach

Euler's Approach

Euler's original "derivation" of the value is clever and original. He essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series. Of course, Euler's original reasoning requires justification, but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the Taylor series expansion of the sine function

Dividing through by x, we have

Now, the roots (intersections of the x-axis) of sin(x)/x occur precisely at where Let us assume we can express this infinite series as a (normalized) product of linear factors given by its roots, just as we do for finite polynomials:

$\begin{align} \frac{\sin(x)}{x} & {} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ & {} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align}$

If we formally multiply out this product and collect all the x2 terms (we are allowed to do so because of Newton's identities), we see that the x2 coefficient of sin(x)/x is

$-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

But from the original infinite series expansion of sin(x)/x, the coefficient of x2 is −1/(3!) = −1/6. These two coefficients must be equal; thus,

$-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

Multiplying through both sides of this equation by gives the sum of the reciprocals of the positive square integers.

$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$

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