A summation that starts from $-\infty$ I need to evaluate $\large\sum_{i=-\infty}^{n} 2^i$.
This is what I tried:
$$S = 2^{-\infty} + \color{red}{2^{-\infty + 1} + 2^{-\infty + 2} + ... + 2^{n-1} + 2^{n}}$$
$$2S = \color{red}{2^{-\infty + 1} + 2^{-\infty + 2} + 2^{-\infty + 3} + ... + 2^{n}} + 2^{n+1}$$
$$
\begin{align*}
2S - S & = (A + 2^{n+1}) - (2^{-\infty} + A)\\
& = 2^{n+1} - 2^{-\infty}
\end{align*}$$
Is my process (and answer) correct?

Let's re-index things to put them in terms of more familiar series.
By letting $j = -i$, this expression is the same as
$$
\sum_{j = -n}^\infty 2^{-j}
$$
If we then let $k = j + n$, then
$$
\sum_{k = 0}^\infty 2^{-k + n} = 2^{n} \sum_{k=0}^\infty 2^{-k}
$$
so we're just left computing $\sum_{k=0}^\infty 2^{-k}$ (which converges by e.g. the ratio test). This is a common geometric series, but if you haven't encountered it before, you can determine its value using the same approach as in your original question:
$$
S = 1 + \frac12 + \frac14 + \dots \\
2S = 2 + 1 + \frac12 + \dots \\
S = 2S - S = 2 + 1 - 1 + \frac12 - \frac12 + \dots = 2
$$
so 
$$
2^{n} \sum_{k=0}^\infty 2^{-k} = 2^n\cdot 2 = 2^{n+1}.
$$
