When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty}  \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty}  \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty}  \left(\frac{x}{2}\right)^n$, does it become $\sum\limits_{n=2}^{\infty}  \left(\frac{nx^{n-1}}{2^n}\right)$? (from $n= 1$, to $n= 2$?) 
If this is how it works, 
is the above equation false? and how could we solve $\sum\limits_{n=1}^{\infty}  \left(\frac{n^2x^n}{2^n}\right)$?
 A: When in doubt, write it out.  You don't need to memorize formulas for how the limits change in differentiation of series.  There are too many cases to consider anyway.  Just look at the first few terms to match the lowest index.
For instance:
$$
\sum_{n=0}^{\infty}  \left(\frac{x}{2}\right)^n
= \sum_{n=0}^{\infty}  \frac{x^n}{2^n}
= 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \cdots
$$
Differentiating the series is going to “kill” the $n=0$ term.  So if we differentiate term-by-term, the first nonzero term is $n=1$:
$$
    \frac{d}{dx}\sum_{n=0}^{\infty}  \frac{x^n}{2^n}
    = \sum_{n=1}^{\infty}  \frac{nx^{n-1}}{2^n}
    = \frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
$$
With the first few terms written out, we can see that this series can also be written as
$$
    \frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
    = \sum_{n=0}^\infty \frac{(n+1)x^n}{2^{n+1}}
$$
This re-indexing trick comes in handy when combining power series: If you can make the exponents match, you can combine like terms.
But in contrast:
$$
    \sum_{n=1}^{\infty}  \frac{x^n}{2^n}
    = \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \cdots
$$
Differentiating this series does not kill off the first term since it's not constant.
Instead,
$$
    \frac{d}{dx}\sum_{n=1}^{\infty}  \frac{x^n}{2^n}
    = \sum_{n=1}^{\infty}  \frac{nx^{n-1}}{2^n}
    = \frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
$$
So there's no change to the lower limit.
A: When differentiating (and I'm following the form you give in the title of he question where the sum stars at he zeroth element, but the principle is he same if you start at the first) $$S(x)=a_0+a_1x+a_2x^2+\dots +a_nx^n +\dots$$ you obtain $$S'(x)=0+a_1+2a_2+3a_3x^3+\dots +(n-1)a_nx^{n-1}+\dots$$
So the term with $a_n$ associates with $x^{n-1}$ (in your case $a_n=2^{-n}$). Note how the zeroth term of the sum for the derivative is now zero, and can be dropped if convenient without changing any of the equations.
You can see that $S(x)=\sum _{n=0}^{\infty}a_nx^n$ and $S'(x)=\sum _{n=0}^{\infty}na_nx^{n-1}=\sum _{n=1}^{\infty}na_nx^{n-1}$ because the zeroth term of the first version of the second sum is zero.
If you want to obtain terms in $n^2$ one way to do it is to note that $xS'(x)=\sum _{n=0}^{\infty}na_nx^{n}$ so that $\left(xS'(x)\right)'=\sum _{n=0}^{\infty}n^2a_nx^{n-1}$
A: No, because the reason the sum lower bound goes from 0 to 1 is because the derivative kills the first term in the sum. When we differentiate the sum starting from $n=1$, the first term is not killed by the derivative and hence you shouldn't drop it from the sum.
To deal with the sum 
$$
F(x)=\sum_{n=1}^\infty n^2 x^n 2^{-n},
$$
Do the following:
Define
$$
S(x)=\frac2{2-x}=\sum_{n=0}^\infty\left(\frac x2\right)^n.
$$
We have shown
$$
S'(x)=\sum_{n=1}^\infty n\left(\frac x2\right)^n=\frac2{(2-x)^2}.
$$
Similarly, we can show
$$
S''(x)=\sum_{n=2}^\infty n(n-1)\left(\frac x2\right)^n=\frac4{(2-x)^3}.
$$
Finally, note that
\begin{equation}\begin{aligned}
F(x)&=\frac x2+\left(\frac x2\right)^2\sum_{n=2}^\infty n(n-1)\left(\frac{x}2\right)^{n-2}+\frac x 2\sum_{n=2}^\infty n\left(\frac x2\right)^{n-1}\\
&=\frac x2+\left(\frac x2\right)^2S''(x)+\frac x2S(x)-\frac x2\\
&=\left(\frac x2\right)^2\frac4{(2-x)^3}+\frac x2\frac2{(2-x)^2}=\frac{4x}{(2-x)^2}.
\end{aligned}\end{equation}
