For an irrational $\alpha>1$, define the sequence $\alpha_n$ such that $\alpha_0=\alpha$ and for $n\geq 1$, $\alpha_n=\frac{1}{\alpha_{n-1}-\lfloor\alpha_{n-1}\rfloor}$. We can also define the sequence $a_n$ so that $a_n=\lfloor\alpha_n\rfloor$ for all $n\geq 0$ (i.e. $[a_0;a_1,a_2,\ldots]$ is the regular continued fraction representation of $\alpha$
Using the notation that the empty product is $1$, prove the following three sums (there's also the issue of convergence) $$\alpha=\sum_{n=0}^\infty\frac{a_{2n}}{\prod_{i=1}^{2n}\alpha_i}$$ $$1=\sum_{n=0}^\infty\frac{a_{2n+1}}{\prod_{i=1}^{2n+1}\alpha_i}$$ $$\alpha=\sum_{n=0}^\infty\frac{a_n}{\prod_{i=1}^n\alpha_i^2}$$
I discovered these results while solving a geometric problem involving starting with a $1\times\alpha$ rectangle and then cutting out squares (similar to how one might draw a Fibonacci spiral in a golden rectangle). All these sums can be seen geometrically from this construction. If we consider side lengths of the squares, we can form an orthogonal path from one corner of the rectangle to the opposite corner and that gives us the first two series (the first is summing the lengths of all the horizontal segments of the path, the second is all the vertical segments). The third series comes from summing the areas of the squares, which of course should have area of the original rectangle.
I'm looking for an analytical proof of these sums. I have verified that these work analytically when $\alpha=\sqrt{2}$ and when $\alpha=\phi$ (both are just simple geometric series). I have no idea how one would go about the general case. I suspect maybe some clever telescoping, but I'm not seeing one.
I also wrote this python code that computes partial sums of the first series (and can be modified easily to do the other two). I wasn't seeing a pattern from this.
import math
a = math.sqrt(2)
n = 20
L = [a]
for i in range(n):
a = L[-1]
L += [1/(a-math.floor(a))]
M = [math.floor(x) for x in L]
s = 0
p = 1
for i in range(n):
if (i%2 == 0):
s += M[i]/p
print(s)
p *= L[i+1]
Using the comment from Greg Martin, I believe a geometric argument leads to concluding that the partial sums evaluate to $$\sum_{n=0}^N \frac{a_{2n}}{\prod_{i=1}^{2n}\alpha_i}=\alpha-\frac{1}{\prod_{i=1}^{2N+1}\alpha_i}$$ $$\sum_{n=0}^N \frac{a_{2n+1}}{\prod_{i=1}^{2n+1}\alpha_i}=1-\frac{1}{\prod_{i=1}^{2N+2}\alpha_i}$$ $$\sum_{n=0}^N \frac{a_n}{\prod_{i=1}^n\alpha_i^2}=\alpha-\frac{1}{\alpha_{N+1}\prod_{i=1}^N\alpha_i^2}$$
So using the comment from Greg Martin and deducing an expression for the partial sums from the geometrical interpretation, I think I have finagled some telescoping manipulations. This is just reverse engineering from the result, which I think is more intuitively discovered from the geometrical interpretation.
The first two results follow from the following. Consider $\frac{a_n}{\prod_{i=1}^n\alpha_i}$. Note that we have \begin{align*} \frac{a_n}{\prod_{i=1}^n\alpha_i}&=\frac{\alpha_n-\frac{1}{\alpha_{n+1}}}{\prod_{i=1}^n\alpha_i}\\ &=\frac{1}{\prod_{i=1}^{n-1}\alpha_i}-\frac{1}{\prod_{i=1}^{n+1}\alpha_i} \end{align*} So we can see that when we sum every other term in the sequence $\frac{a_n}{\prod_{i=1}^n\alpha_i}$ we will get a telescoping.
For the last sum, using the same exact identity, we see \begin{align*} \frac{a_n}{\prod_{i=1}^n\alpha_i^2}&=\frac{\alpha_n-\frac{1}{\alpha_{n+1}}}{\prod_{i=1}^n\alpha_i^2}\\ &=\frac{1}{\alpha_n\prod_{i=1}^{n-1}\alpha_i^2}-\frac{1}{\alpha_{n+1}\prod_{i=1}^n\alpha_i^2} \end{align*} which, again, will telescope.