A Swiss Sum

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 $\dfrac{\pi^2}{6}$.