What is Basel’s problem

Pietro Mengoli, an talian mathematician and clergyman (1626–1686) posed the following problem in 1644:

Find the numerical value of:

\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} +.... = \sum_{k= 1}^{\infty} 1/k^2\]

The problem remained unsolved till 1735 when Euler put forward his brilliant solution. Since he lived in Basel, this problem came to be known as Basel’s problem. He found that its value is $\frac{\pi^2}{6}$. Brilliance of this solution can be understood from the fact that closed form solution of $\sum_{k= 1}^{\infty} 1/k^3$ is still elusive. Here we shall see couple of proofs. There are several other ways to arrive at the result.

Proof 1: Euler’s and first proof of the world, 1735

1. Write $ sinx $ as an infinite product of linear factors

$ sinx = t(1-\frac{t}{\pi})(1+\frac{t}{\pi})(1-\frac{t}{2\pi})(1+\frac{t}{2\pi})… \tag{0} $

2. Begin with $ \sin(\pi x) $ from (0)

\(\begin{align} \sin(\pi x) =& \pi x(1-x)(1+x)(1-\frac{x}{2})(1+\frac{x}{2})... \\\ =& \pi x(1-x^2)(1-\frac{x^2}{4})(1-\frac{x^2}{9})...\\\ =&\pi x\left(1-x^2\left[1 + \frac{1}{4} + \frac{1}{9} + ... \right] - x^4\left[1 + \frac{1}{4.9} + ...\right] - ...\right) \\\ =&\pi x - \pi x^3\left[1 + \frac{1}{4} + \frac{1}{9} + ... \right] - ...\tag{1} \end{align}\) From Taylor’s expansion, \(\begin{align} \sin(\pi x) =& \pi x - \frac{(\pi x)^3}{3!} - \frac{(\pi x)^5}{5!}... \\\ =&\pi x - \pi x^3 .\frac{\pi^2}{6} - \frac{(\pi x)^5}{5!}...\tag{2} \end{align}\) Matching cooefficients of $x^3$ in (1) and (2) we get the required result.

Proof 2: using L’Hospital’s rule

1. Find value of $sin({\pi}y)$

\(\begin{align} sin({\pi}y) &= {\pi}y(1-y)(1+y)(\frac{2-y}{2})(\frac{2+y}{2})...\\\ &= {\pi}y(1-y^2)(\frac{4-y^2}{4})(\frac{9-y^2}{9})... \end{align}\)

2. Take log

\(\begin{align} \ln(sin({\pi}y)) &= \ln(\pi) + \ln(y) + \ln(1-y^2) + \ln(4-y^2) - \ln4 + \ln(9-y^2) - \ln9... \end{align}\)

3. Differentiate with respect to y

\(\frac{\pi cos({\pi}y)}{sin(\pi y)} = \frac{1}{y} - \frac{2y}{1-y^2} - \frac{2y}{4-y^2} - \frac{2y}{9-y^2} - ...\\ \frac{1}{2y^2} - \frac{\pi cos({\pi}y)}{2y \sin(\pi y)} = \frac{1}{1-y^2} + \frac{1}{4-y^2} + \frac{1}{9-y^2} + ...\)

4. $ y = -ix \Longleftrightarrow y^2 = -x^2 $

\(-\frac{1}{2x^2} + \frac{\pi cos(-i{\pi}x)}{2ix \sin(-i\pi x)} = \frac{1}{1+x^2} + \frac{1}{4+x^2} + \frac{1}{9+x^2} + ... \tag{3}\)

5. Use euler’s formula to simplify left hand side

\(\begin{align} \frac{cos(z)}{sin(z)} =& \frac {\frac{1}{2}(e^{iz} + e^{-iz})}{\frac{1}{2i}(e^{iz} - e^{-iz})} = \frac{i(e^{2iz} + 1)}{e^{2iz}-1}&\\\ \frac{\pi cos(-i{\pi}x)}{2ix \sin(-i\pi x)} =& \frac{\pi}{2ix}.\frac{i(e^{2\pi x} + 1)}{(e^{2\pi x}-1)} = \frac{\pi}{2x}.\frac{(e^{2\pi x} - 1) + 2}{(e^{2\pi x}-1)}&\\\ =&\frac{\pi}{2x} + \frac{\pi}{x(e^{2\pi x} - 1)} \end{align}\) \(-\frac{1}{2x^2} + \frac{\pi cos(-i{\pi}x)}{2ix \sin(-i\pi x)} = -\frac{1}{2x^2} + \frac{\pi}{2x} + \frac{\pi}{x(e^{2\pi x} - 1)} = \frac{-e^{2\pi x} + 1 + \pi x(e^{2\pi x} - 1) + 2\pi x }{2x^2(e^{2\pi x} - 1)}\\\ = \frac{-e^{2\pi x} + 1 + \pi xe^{2\pi x} + \pi x }{2x^2(e^{2\pi x} - 1)}\\\)

6. Find value at x = 0 by applying L’Hospital rule

\(\begin{align} \frac{-e^{2\pi x} + 1 + \pi xe^{2\pi x} + \pi x }{2x^2(e^{2\pi x} - 1)} \mapsto& \frac{-2\pi e^{2\pi x} +2\pi^2xe^{2\pi x} + \pi e^{2 \pi x} + \pi}{4x(e^{2\pi x} - 1) +4\pi x^2e^{2\pi x}} \\\ =& \frac{-\pi e^{2\pi x} +2\pi^2xe^{2\pi x} + \pi}{4x(e^{2\pi x} - 1) +4\pi x^2e^{2\pi x}}\\\ =&\frac{\pi}{4}.\frac{-e^{2\pi x} +2\pi xe^{2\pi x} + 1}{x(e^{2\pi x} - 1) +\pi x^2e^{2\pi x}}\\\ \mapsto& \frac{ \, \pi^{3} x e^{\left(2 \, \pi x\right)}}{2 \, \pi^{2} x^{2} e^{\left(2 \, \pi x\right)} + 4 \, \pi x e^{\left(2 \, \pi x\right)} + e^{\left(2 \, \pi x\right)} - 1}\\\ \mapsto& \pi^3.\frac{e^{2\pi x}(2\pi x+1)}{2\pi^2e^{2\pi x}()2x+2\pi x^2) + 4\pi e^{2\pi x}(1+2\pi x}\\\ \mapsto& \frac{\pi^3}{4\pi + 2\pi} = \frac{\pi^2}{6} \end{align}\) which is left hand side of (3) Right hand side of (3) is $ 1 + \frac{1}{4} + \frac{1}{9} + = \sum_1^{\infty} \frac{1}{k^2} $

Hence proved.