Limit of sum of areas of infinite amount of triangles I apologize for the possible incorrect use of math terms since English is not my native language and I'm not a mathematician, but this issue came to my mind about a month ago and I was unable to solve it, so I will appreciate any help.
Let length of a line segment $L$ be $1$. Also define variable $r$ (ratio) that can be any real number on the interval $(0;1)$.
Let us put the vertical line segment with length $L$ starting from the point $(0;0)$ on the orthogonal coordinate system; the other point of this line segment is $(0;L)$. Put the next line segment with the following rules:


*

*Starting point should be located on the X axis, let us assume it as a point $(X(n);0)$, where $X(n)>X(n-1)$;

*Lets treat the previous line segment like a vector, multiply it by $r$. The end point of this vector is the end point of the new line segment.


Here is the example which displays $n = 6$ triangles built with the $L = 1$ and $r = 0.8$.

Initially I tried to solve the following tasks:


*

*Find the function $f(L, r, n)$ which will return the sum of the areas of $n$ triangles giving the length of the line segment is $L$ and a ratio is $r$;

*Define the limit of the $f(L, r, n)$ with $L = 1$; $r \to 1$ and $n \to \infty$;

*Assuming that these issues were solved before, what is the correct way to call this task and where I can find the information about it?



Here is what I have discovered so far.
First, let us find one of the angles of the $n$-th triangle. Define $\beta(n)$ as an angle between $(n-1)$-th and $n$-th line segment; $\alpha(n)$ as an angle between X axis and the $n$-th line segment. For the sake of simplicity let us use $\alpha_N$ and $\beta_N$ instead of a function form.

Since the first triangle is the right triangle, $\sin(\alpha_1)$ is defined by the known relations:
$$\sin(\alpha_1) = \frac {r * L}{L} = r$$
Using the law of sines, investigate the second triangle.
$$\frac {\sin(\alpha_2)}{L*r} = \frac{\sin(\pi - alpha(1))}{L};$$
$$\sin(\alpha_2) = \sin(\pi - \alpha_1) * r = \sin(\alpha_1) * r$$
Since there is no dependency from the right triangle in this formula, we can generalize the result:
$$\sin(\alpha_n) = \sin(\alpha_{n-1})*r$$
or for the calculation simplicity sake:
$$\alpha_n = \arcsin(r^n)$$
Knowing that the sum of the angles of the triangle is $\pi$, we get the following:
$$\pi = \alpha_n + \beta_n + (\pi - \alpha_{n-1});$$
$$0 = \alpha_n + \beta_n - \alpha_{n-1};$$
$$\beta_n = \alpha_{n-1} - \alpha_n$$
Find the area of the $n$-th triangle with the following formula:
$$S(n) = \frac 1 2 * L * L* r * \sin(\beta_n) = \frac 1 2 * L^2 * r * \sin(\beta_n)$$
Such formula is acceptable for the calculations, but we can represent it in a different way.
$$\sin(\beta_n) = \sin(\alpha_{n-1} - \alpha_n) = \sin\alpha_{n-1}*\cos\alpha_n - \sin\alpha_n*\cos\alpha_{n-1}=$$ $$ = r^{n-1}*\sqrt{1-r^{2n}} - r^n*\sqrt{1-r^{2n-2}} $$
Since that is the solution for the first question, I had tried to solve the second, but with no avail.
$$Sum(L,r) = \frac 1 2 *L^2 \lim_{r \to 1, n \to \infty} (r * \sum_{n=1}^\infty \sin(\beta_n)) $$
I also had tried to change the way of area calculation to the sum of the trapezoids, but it wasn't successful as well.
$$S_t(n) = \frac {L*r^{n-1} + L*r^n} 2 * \cos\alpha_n*(1-r)*L$$
I was unable to apply any known to me technique (such as L'Hôpital's rule, Taylor series investigation) to reach the solution, so I resorted to approximate solution.
I have managed to calculate the result of the function $f$ with the $L = 1$; $r = 0.999999$; $n = 100000000$:
$$f(1,0.999999,10^8) = 0.7853973776669734$$
The length of the whole construct was approximately equal to $100.6931$, knowing that the side of the $n$-th triangle on the X axis is:
$$B(n) = L*r*(\frac {\cos\alpha_n} r - \cos\alpha_{n-1}))$$
The result is more or less close to the $\frac \pi 4 (0.7853981633974483)$, which was surprising. I had tried to apply this knowledge (the infinite sum of $\sin\beta_n$ should approach to $\frac \pi 2$; knowledge that $\int_0^{+\infty}\frac {dx}{(1+x)*\sqrt{x}} = \pi$), but was unable to do so.
So here is the final composite question:


*

*Are there any errors in my calculations?

*Is there a way to solve this without resorting to approximations?

*Is this sum really approaches to the $\frac \pi 4$?

*How can I define the curve which is shaped by these triangles?

*Assuming that these issues were solved before, what is the correct way to call this task and where I can find the information about it?


Thank you in advance!
 A: Every point on the limiting curve has a distance 1 to the $x$-axis along the tangent line at that point. Intuitively, this should be plausible. 
More formally, let $x=f(y)$ be the equation for the curve. The equation for the tangent line at $(f(y_0),y_0)$ is
$$x-f(y_0)=f'(y_0)(y-y_0).$$
This meets the $x$-axis at $(-y_0f'(y_0)+f(y_0),0)$.
The condition requires that the distance from $(f(y_0),y_0)$ to $(-y_0f'(y_0)+f(y_0),0)$ equals 1, i.e.:
$$\sqrt{[f(y_0)-(-y_0f'(y_0)+f(y_0))]^2+(y_0-0)^2}=y_0\sqrt{f'(y_0)^2+1}=1$$
$$f'(y)=-\frac{\sqrt{1-y^2}}{y}\tag1,$$
where the minus sign comes from the fact that $x$ decreases as $y$ increases.
Your construction is actually a numerical approximation to the solution of this differential equation, where at every step, you move a distance $1-r$ in the direction given by Eq. (1) evaluated at the last value of $y$. To see this, let the coordinate of the $n^\text{th}$ point $P_n$ be $(x_n,y_n)$. The two points on the $x$-axis that is a distance 1 away from $P_n$ are $(x_n\pm\sqrt{1-y_n^2},0)$, as you can verify. Your construction takes the positive sign. The line from $P_n$ to $(x_n+\sqrt{1-y_n^2},0)$ has equation
$$x-x_n=-\frac{\sqrt{1-y_n^2}}{y_n}(y-y_n)$$
which is exactly the one given by Eq. (1). (I don't yet have a formal proof of its convergence, but I believe it's possible.)
Now, Eq. (1) can be solved by an integration using the substitution $u=\sqrt{1-y^2}$. The result is
$$f(y)=-\sqrt{1-y^2}+\frac{1}{2}\ln\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}.$$
The area can then be calculated by
$\int_0^1f(y)dy$, which can be shown to be equal to $\pi/4$ (again by a substitution $u=\sqrt{1-y^2}$).
A: One can also express the area (as noted in the question) as a sum of trapezoids, obtained by dropping perpendicular lines from $E$, $G$, $I$, $K$, ... to the $x$-axis. The vertical bases of $k$-th trapezoid measure $r^k$ and $r^{k+1}$, while its height is $(1-r)\sqrt{1-r^{2k}}$. Hence total area $S$ can also be computed from:
$$
S=\sum_{k=1}^\infty{1\over2}(r^k+r^{k+1})(1-r)\sqrt{1-r^{2k}}=
{1\over2}(1-r^2)\sum_{k=0}^\infty r^k\sqrt{1-r^{2k}}
$$
(we can start the last sum with $k=0$ because the $0$-th term vanishes).
I tried to get a closed expression for this series with Mathematica, to no avail. We can however use Taylor's expansion $\sqrt{1+x}=\sum_{n=0}^\infty\binom{1/2}{n}x^n$ to write:
$$
S={1\over2}(1-r^2)\sum_{k=0}^\infty r^k\sum_{n=0}^\infty\binom{1/2}{n}(-1)^n r^{2nk}
={1\over2}(1-r^2)\sum_{n=0}^\infty\binom{1/2}{n}(-1)^n\sum_{k=0}^\infty r^{(2n+1)k}
$$
and the last series can be summed (for $r<1$) to give:
$$
S={1\over2}(1-r^2)\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over 1-r^{2n+1}}.
$$
We can now factor and simplify a term $1-r$ both in $1-r^2$ and in $1-r^{2n+1}$:
$$
S={1\over2}(1+r)\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over 1+r+r^2+\dots+r^{2n}}.
$$
The advantage is that we can now carry out the limit $r\to1$ just substituting $r=1$ in this formula, to get:
$$
\lim_{r\to1}S=\sum_{n=0}^\infty\binom{1/2}{n}{(-1)^n\over {2n+1}}.
$$
This is a simpler series and Mathematica evaluates it to $\pi/4$, thus confirming the given conjecture.
EDIT.
With the help of OEIS I found that the above series is Maclaurin expansion of 
$$
x\sqrt{x^2+1} +{1\over2} \ln\big(x+\sqrt{x^2+1}\big)
$$ 
(which is the arc length of Archimedes' spiral), computed for $x=i$ and divided by $i$.
