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 $$|\alpha|<1$$
\(
\begin{align*}
S=\sum_{n=0}^{\infty}\alpha^n&=\frac{1}{1-\alpha}, |\alpha|<1 && \text{Start here} \\
\frac{dS}{d\alpha}=\sum_{n=0}^{\infty}n\alpha^{n-1}&=\frac{1}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{Take derivatives of both sides} \\
\alpha\frac{dS}{d\alpha}=\alpha\sum_{n=0}^{\infty}n\alpha^{n-1}&=\frac{\alpha}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{Multiply both sides by}~\alpha \\
\alpha\frac{dS}{d\alpha}=\sum_{n=0}^{\infty}n\alpha^{n}&=\frac{\alpha}{\left(1-\alpha\right)^2}, |\alpha|<1 && \text{bring}~\alpha~\text{into the summation}
\end{align*}
\)
