Sum of Geometric Sequence

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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

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