Algebraic manipulation of floors and ceilings I am trying to solve the summation
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=3}^{n} \left\lceil\frac{i-2}{2}\right\rceil
$$
I will list some of the simplifications that I've found so far, and then get around to asking my question. Please feel free to point out any logical errors that I have (in all likelihood) made.
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=1}^{n-2} \left\lceil\frac{i}{2}\right\rceil
$$
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\sum_{i=1}^{\left\lceil\frac{n-2}{2}\right\rceil} i+ \sum_{i=1}^{\left\lfloor\frac{n-2}{2}\right\rfloor} i\right)
$$
Since the upper bound in both of the inner summations will be an integer, I have:
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\frac{\left\lceil\frac{n-2}{2}\right\rceil\left(\left\lceil\frac{n-2}{2}\right\rceil + 1\right)}{2}+\frac{\left\lfloor\frac{n-2}{2}\right\rfloor\left(\left\lfloor\frac{n-2}{2}\right\rfloor + 1\right)}{2}\right)
$$
How do I simplify a floor times a floor and a ceiling times a ceiling algebraically?
I'm thinking that the identity
$$
n = \left\lceil\frac{n}{2}\right\rceil + \left\lfloor\frac{n}{2}\right\rfloor
$$
might come in handy, but I have very little experience manipulating floors and ceilings algebraically. Let me be clear: I'm not looking for an answer to the summation, just guidance in simplifying this current step.

\begin{align*}S&=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=3}^n\left\lceil\frac{i-2}{2}\right\rceil&=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=3}^n\left\lfloor\frac{i-1}{2}\right\rfloor\\ &=\sum_{n=3}^\infty\frac{1}{2^n}\sum_{i=2}^{n-1}\left\lfloor\frac{i}{2}\right\rfloor &=\sum_{n=3}^\infty\frac{1}{2^n}\left\lfloor\frac{(n-1)^2}{4}\right\rfloor\end{align*}
with $n=2m\quad$ then $\displaystyle\quad S_{2m}=\sum_{m=2}^\infty\frac{1}{2^{2m}}\left\lfloor\frac{(2m-1)^2}{4}\right\rfloor=\sum_{m=2}^\infty\frac{m^2-m}{2^{2m}}=\sum_{m=1}^\infty\frac{m^2+m}{2^{2m+2}}$
with $n=2m+1\quad$ then $\displaystyle\quad S_{2m+1}=\sum_{m=1}^\infty\frac{1}{2^{2m+1}}\left\lfloor\frac{(2m)^2}{4}\right\rfloor=\sum_{m=1}^\infty\frac{m^2}{2^{2m+1}}$
So $\displaystyle\quad S=S_{2m}+S_{2m+1}=\sum_{m=1}^\infty\frac{3m^2+m}{4^{m+1}}=\boxed{\frac{2}{3}}$
