The Basel Problem

The Basel Problem is a problem in number theory posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1734. As the problem remained open for 90 years, Euler’s solution brought him immediate fame at the age of 28.

The Basel Problem asks the result of the sum

\[ 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+... \]

i.e. the sum of the inverse of the square numbers.

Screen Shot 2016-08-06 at 10.52.38 AM.png

As there are infinitely many square numbers, this sum has infinitely many terms. It is possible to find the ‘result’ of this sum due to the fact that the sum converges to a particular limiting value as the denominator approaches infinity. Part of the difficulty of solving this series is because it converges very slowly; when n = 106, the result is only accurate to 5 decimal places.

Euler found that the result had the value of:

Screen Shot 2016-08-06 at 10.55.45 AM.png

But how did he do this?

Firstly, recall the Taylor series expansion of sin(x):

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

If we divide by x, this gives us:

{\frac {\sin(x)}{x}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+\cdots .

Applying the Weierstrass factorisation theorem, which states that “entire functions can be represented by a product involving their zeroes” we obtain the expression:

{\begin{aligned}{\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{aligned}}

If we multiply out this product and collect all x2 terms we get:

-\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}}}.

In the expansion of sin(x)/x the coefficient of x2 is -1/(3!). By comparing coefficients and noting that they must be equal on each side, we can equate this as:

  -{\frac {1}{6}}=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.

Hence, reaching a final result of

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

Sources: 1 | 2 | 3

M x

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