# Geometric Representations of zeta(2)

In 1735 Euler famously proved that

$\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = 1+ \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \dots = \frac{\pi^2}{6}$

His method was purely algebraic, as are many of the multitude of proofs that appear for this result. However, various geometric perspectives exist, and they may illuminate some of the properties of $\zeta(2)$.

As previously discussed, we may represent $\zeta(2)$ by the double integral

$\displaystyle \zeta (2) = \int_0^1 \int_0^1 \frac{1}{1-xy} \,dx \,dy$

The symmetry of this representation allows us calculate the exact value of $\zeta(2)$. See the previous post for details of this.

Another geometric representation of $\zeta(2)$ is as the area between the curve $e^{-x} + e^{-y} = 1$ and the positive $x$-axis.

Or formulated as an integral

$\displaystyle \int_0^{\infty} - \ln(1-e^{-x}) \,dx = \frac{\pi^2}{6}$

To show that the above integral is indeed equal to $\zeta(2)$ we consider the Mercator series (expansion of the natural logarithm)

$\displaystyle \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots$

so that

\begin{aligned} \displaystyle \int_0^{\infty} - \ln(1-e^{-x}) \,dx & = \int_0^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^n}{n}(-e^{-x})^n \,dx \\ & = \sum_{n=1}^{\infty} \int_0^{\infty} \frac{e^{-nx}}{n} \,dx \\ \end{aligned}

Interchanging the order of summation and integration may be justified as $\displaystyle \frac{e^{-nx}}{n}$ is a sequence of positive integrable functions (see e.g here). What follows is a geometric representation for the individual terms of $\zeta(2)$, through the equation

\begin{aligned} \displaystyle \int_0^{\infty} \frac{e^{-nx}}{n} \,dx & = \Big[ \frac{-e^{-nx}}{n^2} \Big]_0^{\infty} \\ & = \frac{1}{n^2} \end{aligned}

Thus

\begin{aligned} \displaystyle \int_0^{\infty} -\ln(1-e^{-x}) & = \sum_{n=1}^{\infty} \int_0^{\infty} \frac{e^{-nx}}{n} \,dx \\ & = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) \end{aligned}

as originally claimed.

Another way to see that the area under the curve $e^{-x} + e^{-y} = 1$ is $\zeta(2)$ (attributed to  Johan Wästlund) is given in this math.SE post. Indeed most of what follows is tracing through the comments there for my own personal edification.

We construct a geometric object in the positive quadrant of $\mathbb{R}^2$ over $N$ iterations . For example, $D_8$ is shown below. (Image by Hans Lundmark)

We let $D_1$ be the $1 \times 1$ square $[0,1]^2$. Then $D_{N}$ is obtained from $D_{N-1}$ by removing the ‘outermost’ diagonal layer of rectangles and adding two sets of rectangles of area $\frac{1}{N}(\frac{1}{N-1} + \frac{1}{N-2} + \dots + \frac{1}{2} + \frac{1}{1})$ and a square of area $1/N^2$ to cover the ‘inner edges’.

We give a visual representation of this process going from $D_5$ to $D_6$ whereby in constructing $D_6$ from $D_5$ we remove red rectangles and add green ones.

It may be observed that each rectangle removed is equal to two of the rectangles added, namely the $k$th rectangle from the top and the $k+1$th rectangle from the left of those added has area equal to the $k$th rectangle on the diagonal.

\begin{aligned} \frac{1}{N}\frac{1}{k} + \frac{1}{N}\frac{1}{N-k} & = \frac{(N-k)+k}{kN(N-k)} \\ & = \frac{1}{k}\frac{1}{N-k} \end{aligned}

Therefore the area added at stage $N$ in the process above is only that of the square $1/N^2$. Thus the area of $D_N$ is $\sum_{n=1}^N \frac{1}{n^2}$ and thus in the limit the area of $D_N$ approaches $\zeta(2)$.

References

[1] Chapman, R. “Evaluating ζ(2)” 1999. www.secamlocal.ex.ac.uk/~rjc/etc/zeta2.pdf
[2] Passare, M. “How to compute $\sum 1/n^2$ by solving triangles. Amer. Math. Monthly 115, no. 8, 745–752, 2009.