I have the following sum of geometric series:
$$S_{n} = \sum_{k = m}^{n} x^k$$
I want to derive formula for its sum. Derivation:
$$S_{n} = x^m + x^{m+1} + \cdots + x^{n-1} + x^{n}$$
$$S_{n}= x^m \underbrace{\Big(1 + x+\cdots + x^{n-m-1} + x^{n-m}\Big)}_{S'_{n}}$$
$$S_{n} = x^m S'_{n}$$
Let's compute $S'_{n}$ using perturbation method.
$$S'_{n} = \sum_{k=0}^{n-m}x^k$$
$$S'_{n} + x^{n-m+1} = x^0 + \sum_{k=0}^{n-m}x^{k+1}$$
$$S'_{n} + x^{n-m+1} = 1 + x \underbrace{\sum_{k=0}^{n-m}x^{k}}_{S'_{n}}$$
$$S'_{n} + x^{n-m+1} = 1 + x S'_{n}$$
$$S'_{n} – xS'_{n} = 1 – x^{n-m+1}$$
$$S'_{n} \Big( 1 – x \Big) = 1 – x^{n-m+1}$$
$$S'_{n} = \frac{1 – x^{n-m+1}}{1 – x}$$
Plugging back into formula $S_{n} = x^m S'_{n}$
$$\Longrightarrow S_{n} = x^m \frac{1 – x^{n-m+1}}{1 – x}$$
$$S_{n} = \frac{x^m – x^{n+1}}{1 – x}$$
$$\sum_{k=m}^{n} x^k = \frac{x^m – x^{n+1}}{1 – x}$$
My question: what additional assumptions should I make for this formula to be more "official" and work for general purpose?
I was thinking assumptions like:
- $x \neq 1$
- $m \geq 0 \quad \land \quad n > m$
Not sure if these assumptions make sense. What else should I consider?
Best Answer
This is a good derivation. You don't need to consider anything else than what you have written. Since you don't want to divide through zero, you have to assume $x \neq 1$ and for the summation it is correct to assume $m \geq 0$ and $n > m$.