# A Swiss sum

Can you use the given image to say something about the sum of an infinite series?

## Problem

The Basel problem, dating back to the 17th century, asks for the sum of the infinite series

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots$$

What does this image tell you about the sum?

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots$$

What does this image tell you about the sum?

Image

## Getting Started

What are the sizes of the squares in the diagram?

How far apart are the horizontal lines in the second large square?

How would the diagram continue?

How far apart are the horizontal lines in the second large square?

How would the diagram continue?

## Student Solutions

If the large squares are taken to have side length 1, then the left-hand large square has area $\dfrac{1}{1^2}$.

In the right-hand large square, the blue squares in the top row have side lengths $\dfrac{1}{2}$ and $\dfrac{1}{3}$ and so their areas sum to $\dfrac{1}{2^2}+\dfrac{1}{3^2}$.

The squares in the second row have side lengths $\dfrac{1}{4}$, $\dfrac{1}{5}$, $\dfrac{1}{6}$, $\dfrac{1}{7}$, and so their areas sum to $\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}$.

The squares in the third row have side lengths $\dfrac{1}{8}$, $\dfrac{1}{9}$, ..., $\dfrac{1}{15}$, and so their areas sum to $\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots+\dfrac{1}{15^2}$.

Continuing in this way, there are $2^n$ squares in the $n$th row with side lengths $\dfrac{1}{2^n}$, $\dfrac{1}{2^n+1}$, ..., $\dfrac{1}{2^{n+1}-1}$ and their areas sum to $\dfrac{1}{(2^n)^2}+\dfrac{1}{(2^n+1)^2}+\cdots+\dfrac{1}{(2^{n+1}-1)^2}$. These squares all fit in the row, as $\dfrac{1}{2^n}+\dfrac{1}{2^n+1}+\cdots+\dfrac{1}{2^{n+1}-1}<\dfrac{1}{2^n}+\dfrac{1}{2^n}+\cdots+\dfrac{1}{2^n}=2^n\times \dfrac{1}{2^n}=1$.

Therefore the areas of all of the blue squares sum to $\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots$.

The height of the first row in the right-hand large square is the size of the first square, which is $\dfrac{1}{2}$. The height of the second row is $\dfrac{1}{4}$, of the third row is $\dfrac{1}{8}$, and so on. So the heights of all of the rows sum to $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots=1$. This means that all of the blue squares in the right-hand diagram do fit inside the large square, and the sum of their areas is less than the area of the large square, which is 1. Therefore

$$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots<2.$$

Euler showed that the exact value of this sum is $\dfrac{\pi^2}{6}$.

In the right-hand large square, the blue squares in the top row have side lengths $\dfrac{1}{2}$ and $\dfrac{1}{3}$ and so their areas sum to $\dfrac{1}{2^2}+\dfrac{1}{3^2}$.

The squares in the second row have side lengths $\dfrac{1}{4}$, $\dfrac{1}{5}$, $\dfrac{1}{6}$, $\dfrac{1}{7}$, and so their areas sum to $\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}$.

The squares in the third row have side lengths $\dfrac{1}{8}$, $\dfrac{1}{9}$, ..., $\dfrac{1}{15}$, and so their areas sum to $\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots+\dfrac{1}{15^2}$.

Continuing in this way, there are $2^n$ squares in the $n$th row with side lengths $\dfrac{1}{2^n}$, $\dfrac{1}{2^n+1}$, ..., $\dfrac{1}{2^{n+1}-1}$ and their areas sum to $\dfrac{1}{(2^n)^2}+\dfrac{1}{(2^n+1)^2}+\cdots+\dfrac{1}{(2^{n+1}-1)^2}$. These squares all fit in the row, as $\dfrac{1}{2^n}+\dfrac{1}{2^n+1}+\cdots+\dfrac{1}{2^{n+1}-1}<\dfrac{1}{2^n}+\dfrac{1}{2^n}+\cdots+\dfrac{1}{2^n}=2^n\times \dfrac{1}{2^n}=1$.

Therefore the areas of all of the blue squares sum to $\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots$.

The height of the first row in the right-hand large square is the size of the first square, which is $\dfrac{1}{2}$. The height of the second row is $\dfrac{1}{4}$, of the third row is $\dfrac{1}{8}$, and so on. So the heights of all of the rows sum to $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots=1$. This means that all of the blue squares in the right-hand diagram do fit inside the large square, and the sum of their areas is less than the area of the large square, which is 1. Therefore

$$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots<2.$$

Euler showed that the exact value of this sum is $\dfrac{\pi^2}{6}$.

## Teachers' Resources

### Why do this problem?

After geometric series, this is one of the simplest infinite series with finite sum, all of whose terms are positive. This geometric demonstration of the result requires students to continue a pattern and to use several steps of reasoning to deduce that the sum is bounded by 2. Summing infinite geometric series also play an important role in the this proof, so this could be used to show an application of them in a larger proof. (It would be useful for students to be able to sum $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ before tackling this problem.)

The exact value of this sum was first discovered by Euler after having defeated many other great mathematicians of the time: it is $\frac{\pi^2}{6}$.

### Possible approach

Students might consider the problem Harmonically before or after this problem, which shows a start contrast between similar-looking series, one of which converges and the other diverges.

For this problem, students could be asked to make a prediction about the sum of the series before being shown the image. They could then be given the image and asked to work out how it relates to the sum, and what they can deduce from it.

A more challenging variant would be to ask students if they can show geometrically that the series sums to less than 2 without giving them an image to work with.

### Key questions

What are the side lengths of the blue squares?

How does the image continue?

Will all of the necessary blue squares fit in the two large squares?

### Possible extension

What can you say about the sum of the series $\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\cdots$ if $s$ is a number other than 2?

### Possible support

Students might need help identifying what they need to justify: in particular, they may not show that the squares fit in each row, and that all the rows fit inside the right-hand square.

### Further exploration

One of the assignments on the STEP Support Programme, Assignment 24, has a warm-down problem which explores this sum a little further. A video showing a beautiful geometric argument has been made by 3Blue1Brown and is available here. Many other approaches to calculating the exact sum can be found online.