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?
 A: 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 
$$
2^{n} \sum_{k=0}^\infty 2^{-k} = 2^n\cdot 2 = 2^{n+1}.
$$
A: Here is an intuitive explanation; to turn it into a rigourous proof, see below.
Let $$S = 2^n + 2^{n-1} + \cdots$$Then $$\begin{align}2S &= 2^{n+1}+2^n+2^{n-1}+\cdots\\ &= 2^{n+1} + S\end{align}$$
So $S = 2S-S = 2^{n+1}$
Note that in doing this, we don't have to think about what "$2^{-\infty}$" would be; this value isn't strictly defined. This is an infinite series, and as such it has no final term. Given any term, I can create the next term from it by dividing by 2, so if there were a final term, then I could divide by 2 to get another term!

In order to prove this rigourously we must make sure that the series converges, so that $S$ actually has a value that we can manipulate arithmetically. Define $$S_k=\sum_{i=-k}^n2^i$$
Then $$\begin{align}
2S_k&= \sum_{i=-k}^n2^{i+1}\\
&=\sum_{i=-k}^n2^i + 2^{n+1}-2^{-k}\\
&=S_k + 2^{n+1}-2^{-k}
\end{align}$$
So $$S_k = 2^{n+1} -2^{-k}$$ and as $k \to \infty$, $2^{-k} \to 0$, so $$S_k \to 2^{n+1}$$
A: Notice that
$$\sum_{i=-\infty}^n2^i=\sum_{i=-n}^{\infty}2^{-i}=\sum_{i=-n}^02^{-i}+\sum_{i=1}^{\infty}2^{-i}=\sum_{i=0}^n2^i+\sum_{i=1}^\infty\frac{1}{2^i}$$
This reduces the problem to show that $\sum_{i=1}^\infty\frac{1}{2^i}=1$ which is a very well documented result and that $\sum_{i=0}^n2^i=2^{n+1}-1$ which can be shown with induction.
A: Such "infinite sums" are not really sums. They are limits of finite sums. You don't compute them by adding up infinitely many terms--it isn't possible to perform an infinite number of steps in a computation. The symbol
$$S = \sum_{i=-\infty}^n 2^i$$
means
$$S = \lim_{N\rightarrow\infty}\sum_{i=-N}^n 2^i$$
This is computed by first finding the value of
$$S_N \equiv \sum_{i=-N}^n 2^i = 2^{-N} +\cdots + 2^n$$
and then computing the limit
$$S = \lim_{N\rightarrow\infty}S_N$$
To find what $S_N$ is, just note that
$$S_N =2S_N -S_N = 2(2^{-N} +\cdots + 2^n)-(2^{-N} +\cdots + 2^n)$$
$$=(2^{-N+1} +\cdots + 2^{n+1}) - (2^{-N} +\cdots + 2^n)$$
$$=2^{n+1} - 2^{-N}$$
Then you can see at once that
$$S = \lim_{N\rightarrow\infty}S_N = \lim_{N\rightarrow\infty}(2^{n+1} - 2^{-N}) = 2^{n+1}$$
