Introduction
This page contains the most common summation formulas as well as short examples of how to obtain them.
The Formulas
The first formula is the generic one for any $ N $ and $ \alpha $. The next two are for cases where $ N=\infty $ and $ |\alpha|<1 $. The fourth has a slightly different summand.
$ \begin{align} \sum_{n=k}^{N-1}\alpha^n&= \left\{ \begin{array}{ll} N-k & \alpha=1\\ \frac{\alpha^k-\alpha^N}{1-\alpha} & \alpha\neq 1 \end{array} \right. \\ \sum_{n=0}^{\infty}\alpha^n&= \frac{1}{1-\alpha}\mbox{ if }|\alpha|<1\\ \sum_{n=k}^{\infty}\alpha^n&= \frac{\alpha^k}{1-\alpha}\mbox{ if }|\alpha|<1 \\ \sum_{n=0}^{\infty}n\alpha^n&= \frac{\alpha}{(1-\alpha)^2}\mbox{ if }|\alpha|<1 \end{align} $
The Derivations
Generic $ \alpha^n $ summand
For the first formula, let's call the sum $ S $:
$ \begin{align*} S&=\sum_{n=k}^{N-1}\alpha^n \end{align*} $
If $ \alpha=1 $, we end up with:
$ \begin{align*} S&=\sum_{n=k}^{N-1}1 \end{align*} $
and since there are $ N-1-k+1=N-k $ terms from $ k $ to $ N-1 $, we add up $ N-k $ 1's to get $ N-k $. Done with that particular case!
If $ \alpha\neq 1 $, let's look at what happens if we multiply $ S $ by $ \alpha $ and then do some index transformations:
$ \begin{align*} S&=\sum_{n=k}^{N-1}\alpha^n && \text{Start here} \\ \alpha S&=\alpha\sum_{n=k}^{N-1}\alpha^n && \text{Multiply both sides by}~\alpha \\ \alpha S&= \sum_{n=k}^{N-1}\alpha^{n+1} && \text{Bring}~\alpha~\text{into summand, which increases power by 1}\\ \alpha S&= \sum_{m=k+1}^{N}\alpha^{m} && \text{Transform indexing}~m=n+1~ \text{so summand is the same as the one for}~S\\ \end{align*} $
Now this has the right summand, but the limits are wrong. This new summation starts at $ k+1 $ instead of $ k $. It ends at $ N $ instead of $ N-1 $. If we start at $ m=k $, we get an extra $ \alpha^{k} $ term in the summation -- we can subtract that from the summation to maintain the original summation's value:
$ \begin{align*} \alpha S&= \sum_{m=k}^{N}\alpha^{m} - \alpha^k && \text{Switch lower index to match and subtract off extra term}\\ \end{align*} $
To fix the upper limit, change it to $ N-1 $ but now the summation is missing the $ \alpha^N $ term, so we need to add that back:
$ \begin{align*} \alpha S&= \sum_{m=k}^{N-1}\alpha^{m} - \alpha^k + \alpha^N && \text{Switch upper index to match and add in the missing term}\\ \end{align*} $
Now look at the difference between $ S $ and $ \alpha S $:
$ \begin{align*} S - \alpha S&= \sum_{n=k}^{N-1}\alpha^n - \left( \sum_{m=k}^{N-1}\alpha^{m} - \alpha^k + \alpha^N \right) \end{align*} $
Even though they have different dummy variable, the two sums are now exactly the same, meaning:
$ \begin{align*} S - \alpha S&= \alpha^k - \alpha^N \\ S&=\frac{\alpha^k - \alpha^N}{1-\alpha}, \alpha\neq 1 \end{align*} $
$ \alpha^n $ summand with $ N=\infty $ if $ |\alpha|<1 $
As long as $ |\alpha|<0 $, $ \lim_{N\rightarrow\infty}\alpha^N=0 $, so:
$ \begin{align*} \sum_{m=k}^{\infty}\alpha^{m}&=\frac{\alpha^k}{1-\alpha}, |\alpha|<1 \end{align*} $
$ \alpha^n $ summand with $ N=\infty $ and $ k=0 $ if $ |\alpha|<1 $
As long as $ |\alpha|<0 $, $ \lim_{N\rightarrow\infty}\alpha^N=0 $, and if $ k=0 $, $ \alpha^k=\alpha^0=1 $, so:
$ \begin{align*} \sum_{m=0}^{\infty}\alpha^{m}&=\frac{1}{1-\alpha}, |\alpha|<1 \end{align*} $
$ n\alpha^n $ summand with $ N=\infty $ and $ k=0 $ if $ |\alpha|<1 $
Note - while there is a form of this for generic lower bound $ n=k $ and generic upper bound$ n=N-1 $, the cleanest one is in the case where $ k=0 $ and $ N=\infty $, which is only bounded if $ |\alpha|<1 $
$ \begin{align*} S=\sum_{n=0}^{\infty}\alpha^n&=\frac{1}{1-\alpha}, |\alpha|<1 && \text{Start here} \\ \frac{dS}{d\alpha}=\sum_{n=0}^{\infty}n\alpha^{n-1}&=\frac{1}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{Take derivatives of both sides} \\ \alpha\frac{dS}{d\alpha}=\alpha\sum_{n=0}^{\infty}n\alpha^{n-1}&=\frac{\alpha}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{Multiply both sides by}~\alpha \\ \alpha\frac{dS}{d\alpha}=\sum_{n=0}^{\infty}n\alpha^{n}&=\frac{\alpha}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{bring}~\alpha~\text{into the summation} \end{align*} $