<s>[INST] 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.
 [/INST] Start with the generating function for $\displaystyle a_i=\left\lceil \frac{i-2}{2}\right\rceil\; , i\ge 3$
which turns out to be 
\begin{align*}
  G(x) &= \frac{x^3}{1-x-x^2+x^3}
\end{align*}
and for the sum of the coefficients, it's 
\begin{align*}
  H(x) &= \frac{1}{1-x}\, G(x)
\end{align*}
and the required answer is $\displaystyle H\left(\frac{1}{2}\right)$
</s>