This doesn't seem quite correct. For one thing the final condition does not make sense as written. It should probably be  " It is not the case that $d_{2i}=a_{2i+1}$ (so $d_{2i-1}=0$) for all large enough $i$ " (equivalently "$d_{2i} \lt a_{2i+1}$ infinitely often") .  One would also want the similar condition for the situation $d_{2i-1}=a_{2i}.$ 

Perhaps $d_0$ can be any integer, positive or negative. Otherwise there is certainly an upper bound (perhap$(a_0+1)\theta_1$) on the  $\beta$  which can be expressed. 

HOWEVER, something along these lines should be true. You could likely discover then prove it by (imagining that) you are looking at an example such as $\sqrt{3}=[3;3,3,3,\cdots]$, considering the sums with $d_i=0$ for $i \gt N$ and where they are: The following would seem to  need to be true along with the obvious extensions. Only the indicated inequalities actually need to be justified.  $$0 \lt \theta_0 \lt 2\theta_0 \lt 3\theta_0$$  $$(d_0-1)\theta_0 \lt^? d_0\theta_0+3\theta_1 \lt d_0\theta_0+2\theta_1 \lt d_0\theta_0+\theta_1 \lt d_0\theta_0 $$ 

$$d_1\theta_1 \lt d_1\theta_1+\theta_2 \lt d_1\theta_1+2\theta_2\lt d_1\theta_1+3\theta_2  \lt^? (d_1-1)\theta_1$$

Describe what happens (for general $\alpha$), prove it using the properties of continued fractions, figure out what breaks down with $d_i=a_{i+1}$ and $d_{i-1} \ne 0$ and with having the (correct) final conditions fail.

**Later** A alternating in sign sequence $|\theta_0| \gt |\theta_2| \gt \cdots$  from a continued fraction has many properties but perhaps only some of them are needed for the result you wish (whatever it is). It might be helpful to just analyze what has to be true for the result to hold.