# Assumptions for sum of geometric series' derivation?

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?

• Your derivation is good. You dont need x diferent from 1 . This formula is general and official, not sure ehat you mean – Milan Sep 4 at 18:47
• @Milan Of course OP needs $x \neq 1$ in this formula. – Jan Sep 4 at 18:58
• @Jan oops , my bad 😅 – Milan Sep 4 at 20:31

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 $$n > m$$.