I was surprised this week by a certain infinite series, involving reciprocals of cubes.

I was working through a famous bit of math, while getting ready for a course I'm planning on teaching this summer. This is the Basel Problem: Find the sum of the series \[ \begin{align*}\sum_{n=1}^\infty\frac1{n^2} &= 1 + \frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \frac1{5^2} + \cdots \\ &= 1+\frac14 + \frac19 + \frac1{16} + \frac1{25} + \cdots \end{align*}\] This was solved in 1734 by Leonhard Euler, who showed \[ 1+\frac14 + \frac19 + \frac1{16} + \frac1{25} + \cdots = \frac{\pi^2}{6}. \] Euler also found similar formulas for sums of reciprocals of other even powers, such as fourth powers: \[ 1 + \frac1{2^4} + \frac1{3^4} + \frac1{4^4} + \frac1{5^4} + \cdots = \frac{\pi^4}{90}. \]

Now anyone who teaches calculus is likely aware that no such formula is available for sums of reciprocals of cubes: \[ 1 + \frac1{2^3} + \frac1{3^3} + \frac1{4^3} + \frac1{5^3} + \cdots = \text{???} \] This appears to be non-elementary; it's evidently not a rational multiple of \(\pi^3\). It was only in 1978 that the French mathematician Roger Apéry proved this number is irrational. It is called Apéry's constant in his honor.

Now one way of proving formulas of this sort is to use Fourier series. For example, if we define \(f(x)=x^2\) on the interval \([-\pi,\pi]\), and extend it to be \(2\pi\)-periodic, we can represent the function by a cosine series: \[ f(x) = a_0 + a_1 \cos x + a_2\cos 2x + a_3\cos 3x + \cdots\] The recipe for the coefficients is \[ a_n = \frac1\pi\int_{-\pi}^\pi f(x)\cos nx \, dx.\] for \(n\gt0\), and \[ a_0 = \frac1{2\pi} \int_{-\pi}^\pi f(x)\, dx .\] Here, this becomes \[ a_0 = \frac1{2\pi}\int_{-\pi}^\pi x^2\, dx = \frac{\pi^2}3 \] and \[ a_n = \frac1{\pi}\int_{-\pi}^\pi x^2\cos nx \, dx = \frac{4(-1)^n}{n^2}. \] Thus the function is represented by the series \[ x^2 = \frac{\pi^2}3 + \sum_{n=1}^\infty \frac{4(-1)^n}{n^2}\, \cos nx.\] Now if we let \(x=\pi/2\), we get \[ \frac{\pi^2}4 = \frac{\pi^2}3 + \sum_{k=1}^\infty \frac{4(-1)^k}{(2k)^2} \] or \[ \frac{\pi^2}{12} = 1 - \frac1{2^2} + \frac1{3^2} - \frac1{4^2} + \cdots \] To obtain the series for the Basel Problem itself, substract this series from \[ S = 1 + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \cdots \] to get \[ \begin{align*} S - \frac{\pi^2}{12} &= \frac{2}{2^2} + \frac{2}{4^2} + \frac{2}{6^2} + \cdots \\ &= \frac12\left( 1 + \frac1{1^2} + \frac1{2^2} + \frac1{3^2} + + \cdots \right) = \frac{S}2. \end{align*}\] This solves to give \(S=\pi^2/6\).

Now at this point, we haven't shown these calculations are correct; we would need to prove that the Fourier series for \(x^2\) does indeed converge to that function. This will be one of the key ideas the course I'm building will focus on.

But as I worked on the example I just showed, I wondered what would happen if Fourier series were used to try to find a formula for Apéry's constant. Presumably, something prevents the method from working, else it would yield a simple value for the constant.

The method would involve finding a Fourier series for the function \(f(x)=x^3\). Since this function is odd (in the sense that \(f(-x)=-f(x)\) for all \(x\)), we expect it to have a sine series: \[x^3 = a_1\sin x + a_2\sin 2x + a_3\sin 3x + a_4\sin 4x + \cdots\] Here, the recipe for the \(n\)th coefficient is \[a_n = \frac1\pi\int_{-\pi}^\pi f(x)\sin nx\, dx,\] and this time, three integrations by parts gives \[a_n = \frac1\pi\int_{-\pi}^\pi x^3\sin nx\, dx = (-1)^{n+1}\left(\frac{2\pi^2}n + \frac{12}{n^3}\right)\] So we have \[ x^3 = \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{2\pi^2}n + \frac{12}{n^3}\right)\sin nx\] Now the method to find the sum of the series \(\sum_{n=1}^\infty 1/n^3\) would be to let \(x=\pi/2\). The only complication at the outset is that while we see the needed terms of \(1/n^3\), there are also terms with \(1/n\) in them. However, these are easily dealt with using the familiar series \[ \frac\pi4 = 1 - \frac13 + \frac15 - \frac17 + \frac19 - \cdots\] Now with some simple algebra, we arrive at the following series: \[ \sum_{n=1}^\infty(-1)^{n+1}\frac1{(2n-1)^3} = 1 - \frac1{3^3} + \frac1{5^3} - \frac1{7^3} + \cdots = \frac{\pi^3}{32}.\] This surprised me I did not expect to see any series involving reciprocals of cubes with a simple value. I wondered if I'd made some dimwitted mistake, but except for a factor of \(2\), my calculations were correct. (I originally had the value as \(\pi^3/16\).)

But Apéry's constant is still unresolved. Recall with the cosine series for \(x^2\) we first found \[\frac{\pi^2}{12} = 1 - \frac1{2^2} + \frac1{3^2} - \frac1{4^2} + \cdots\] and then with a little work, we found the desired series for the Basel Problem, \[ 1+\frac14 + \frac19 + \frac1{16} + \frac1{25} + \cdots = \frac{\pi^2}{6}.\] Here, it turns out, is the sticking point for Apéry's constant: this doesn't work if we start with the alternating sum involving odd cubes. We can't add or subtract this series from Apéry's constant to get a hold of that constant. It's curious that the constant remains aloof in this way. But perhaps this phase change is not so surprising. We know that \[ \frac\pi4 = 1 - \frac13 + \frac15 - \frac17 + \frac19 - \cdots \] but a seemingly-similar series has a rather different sum: \[ \ln 2 = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \cdots\] We presume Apéry's constant is likewise rather different, but involving a non-elementary function instead of a logarithm.

But I was surprised that the series with odd cubes is actually elementary.