Series of Squares Under Triangle

Question

A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.

Answer 1

Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $\frac sa$. Within a single instance of "square + hat", the square occupies a share of $\frac{s^2}{s^2+\frac12\cdot s\cdot \frac sa}=\frac{1}{1+\frac1{2a}}=\frac{2a}{2a+1}$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit. We conclude that the total area occupied by the squares is $$ \frac{2a}{2a+1}\cdot \frac a2=\frac{a^2}{2a+1}.$$