Does this always converge? So, I was researching the whole quadrature thing recently and decided to define right triangles area(with legs a and b) as a sum of the inscribed square and two smaller triangles.
Square inscribed in right triangle with legs a and b
To find its area I have defined following recursive expression:
$$S\bigl(a, b\bigr) = \bigl(\frac{ab}{a+b}\bigr)^2 + S\bigl(\frac{a^2}{a+b}, \frac{b}{a+b}\bigr) + S\bigl(\frac{ab}{a+b}, \frac{b^2}{a+b}\bigr)$$
And, as I was trying to find S(a, b) in finite terms(and assuming that I don't know that it is $\frac{ab}{2}$), expanding S on the right side, I came up to following:
$$S\bigl(a, b\bigr) = \bigl(\frac{ab}{a+b}\bigr)^2\sum_{i=0}^\infty \biggl(\frac{a^2}{(a+b)^2}+\frac{b^2}{(a+b)^2}\biggr)^i$$
And thus, $S\bigl(a, b\bigr) = \frac{ab}{2}$
But what if we don't limit this problem to just positive a and b?
Does it converge?
And to what it converges?
Edit: Typo in series

For starters, for $S$ to be well defined we need $a+b \neq 0$. For each $(a,b) \in \mathbb{R}$, $\bar{S}(a,b) := S(a,b)(\frac{a+b}{ab})^2$ is nothing more than a geometric series. It will converge if and only if 
$$
\big|\frac{a^2+b^2}{(a+b)^2}\big| = \frac{a^2+b^2}{(a+b)^2} < 1
$$
This will occur if $a^2 + b^2 < (a+b)^2 = a^2 +2ab + b^2$, that is, when $2ab >0$. Therefore, both $a$ and $b$ have to share the same sign and be nonzero. As for the value of convergence, since we're dealing with a geometric series,
$$
\bar{S}(a,b) = \frac{1}{1-\frac{a^2+b^2}{(a+b)^2}} = \frac{1}{\frac{2ab}{(a+b)^2}} = \frac{(a+b)^2}{2ab} 
$$
Now,
$$
S(a,b) =  \big(\frac{ab}{a+b}\big)^2\bar{S}(a,b) = \frac{a^2b^2}{2ab} = \frac{ab}{2}
$$
like predicted. To sum up, the biggest domain we can consider is $D := \{(a,b) : ab > 0\}$, and here $S(a,b) = \frac{ab}{2}$, coinciding with the values known for positive $a$ and $b$.
