∑1n2 Converges

The sum of the reciprocals of the square numbers was a particularly difficult challenge, first posed around 1650, and later named the Basel problem.

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots$$

Although there are more modern proofs, we will follow Euler’s original proof from 1734 because his methods were pretty audacious, and later influenced Riemann’s work on the prime number theorem.

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Taylor Series For sin(x)

We start with the familiar Taylor series for $\sin (x)$. 

$$\sin (x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\dots$$

Euler’s New Series For sin(x)

The polynomial $f(x)=(1-\frac{x}{a})(1+\frac{x}{a})$ has factors $(1-\frac{x}{a})$ and $(1+\frac{x}{a})$, and zeros at $+a$ and $-a$. We can shorten it to $f(x)=(1-\frac{x^2}{a^2})$.

Euler’s novel idea was to write $\sin (x)$ as a product of similar linear factors, which would lead him to a different series.

The zeros of $\sin (x)$ are at $0,\pm \pi ,\pm 2\pi ,\pm 3\pi ,\dots$ so the product of factors looks like the following.

$$\sin (x)=A\cdot x\cdot (1-\frac{x^2}{{\pi }^2})\cdot (1-\frac{x^2}{(2\pi {)}^2})\cdot (1-\frac{x^2}{(3\pi {)}^2})\cdot \dots$$

The constant $A$ is 1 because we know $\frac{\sin (x)}{x}\to 1$ as $x\to 0$. Alternatively, taking the first derivative of both sides gives $A=1$ when $x=0$.

The second factor is $x$ and not $x^2$ because the zero of $\sin (x)$ at $x=0$ has multiplicity 1.

Euler then expanded out the series.

$$\sin (x)=x\cdot [1-\frac{x^2}{{\pi }^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots )+X]$$

Inside the square brackets, the terms with powers of $x$ higher than 2 are contained in $X$.

Comparing The Two Series

The terms in Euler’s new series and the Taylor series must be equivalent because they both represent $\sin (x)$. Let’s pick out the $x^3$ terms from both series.

$$\frac{x^3}{3!}=\frac{x^3}{{\pi }^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots )$$

We can easily rearrange this to give us the desired infinite sum.

$$\frac{{\pi }^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots$$

Euler, aged 28, had solved the long standing Basel problem, not only proving the infinite series of squared reciprocals converged, but giving it an exact value.

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

Rigour

Euler’s original proof was adventurous in expressing $\sin (x)$ as an infinite product of simple linear factors. It made intuitive sense, but at the time was not rigorously justified.

It was almost 100 years later when Weierstrass developed and proved a factorisation theorem that confirmed Euler’s leap was legitimate.