Sum of Geometric Sequence
Theorem
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
- $\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
Corollary 1
Let $a, a r, a r^2, \ldots, a r^{n - 1}$ be a geometric sequence.
Then:
- $\ds \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {r^n - 1} } {r - 1}$
Corollary 2
- $\ds \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$
Proof 1
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \dfrac {x^1 - 1} {x - 1}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^{1 - 1} x^j\) |
so $\map P 1$ holds.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \sum_{j \mathop = 0}^{k - 1} x^j = \frac {x^k - 1} {x - 1}$
Then we need to show:
- $\ds \sum_{j \mathop = 0}^k x^j = \frac {x^{k + 1} - 1} {x - 1}$
Induction Step
This is our induction step:
\(\ds \sum_{j \mathop = 0}^k x^j\) | \(=\) | \(\ds \sum_{j \mathop = 0}^{k - 1} x^j + x^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^k - 1} {x - 1} + x^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^k - 1 + \paren {x - 1} x^k} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^k - 1 + x^{k + 1} - x^k} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{k + 1} - 1} {x - 1}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \N_{>0}: \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
$\blacksquare$
Proof 2
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.
Then:
\(\ds \paren {x - 1} S_n\) | \(=\) | \(\ds x S_n - S_n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^n + \sum_{j \mathop = 1}^{n - 1} x^j - \paren {x^0 + \sum_{j \mathop = 1}^{n - 1} x^j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^n - x^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^n - 1\) |
The result follows.
$\blacksquare$
Proof 3
From Difference of Two Powers:
- $\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Set $a = x$ and $b = 1$:
- $\ds x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = \paren {x - 1} \sum_{j \mathop = 0}^{n - 1} x^j$
from which the result follows directly.
$\blacksquare$
Proof 4
Lemma
Let $n \in \N_{>0}$.
Then:
- $\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
$\Box$
Then by the lemma:
\(\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i\) | \(=\) | \(\ds 1 - x^n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}\) | \(=\) | \(\ds \frac {x^n - 1} {x - 1}\) |
$\blacksquare$
Also presented as
Note that when $x < 1$ the result is usually given as:
- $\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {1 - x^n} {1 - x}$
Some sources give it as:
- $\ds \sum_{j \mathop = 0}^n x^j = \frac {1 - x^{n + 1} } {1 - x}$
and likewise its corollary:
- $\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
Examples
$\dfrac 1 7$ from $1$ to $n$
- $\ds \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \paren {1 - \frac 1 {7^{n + 1} } }$
Common Ratio $1$
Consider the Sum of Geometric Sequence defined on the standard number fields for all $x \ne 1$.
- $\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
- $\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$
Index to $-1$
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
- $\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
- $\ds \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$
Index to $-2$
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
- $\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
- $\ds \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$
Also see
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Theorem $\text{1-2}$
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.1$: The Geometric Series
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.5$: Fermat's Calculation of $\int_0^b x^n \rd x$ for Positive Rational $n$