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Rule

Sum of a Geometric Series

Let

a_{1}

and

r

be the first term and the common ratio, respectively, of a geometric sequence with

n

terms, where

r \neq = 1 .

The sum of the related finite geometric series or the partial sum of the infinite series can be found by using the following formula.

$S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r},$ $r \neq = 1$

For an infinite series, if the common ratio $r$ is greater than $- 1$ and less than $1$ — in other words, if $∣ r ∣ < 1$ — then the sum can be found by using the following formula.

$S_{\infty} = \frac{a _{1}}{1 - r},$ $- 1 < r < 1$

This means that the sum converges on a number. If the common ratio $r$ is less than or equal to $- 1$ or greater than or equal to $1$ — if $∣ r ∣ \geq 1$ — then the sum diverges. In such cases, there is no sum for the infinite geometric series.

Proof

Finite Sum

To derive the formula for a geometric series, a geometric sequence of

n

terms with the first term

a_{1}

and the common ratio

r

will be considered.

a_{1}, a_{1} r, a_{1} r^{2}, \dots, a_{1} r^{n - 1}

All the terms will be added to express the series and

a_{1}

will be factored out.

S_{n} = a_{1} + a_{1} r + a_{1} r^{2} + \dots + a_{1} r^{n - 1}

Factor out $a_{1}$

S_{n} = a_{1} (1 + r + r^{2} + \dots + r^{n - 1})

Then, both sides of the equation will be multiplied by

(1 - r)

and the resulting equation simplified.

S_{n} = a_{1} (1 + r + r^{2} + \dots + r^{n - 1})

$LHS \cdot (1 - r) = RHS \cdot (1 - r)$

S_{n} (1 - r) = a_{1} (1 + r + r^{2} + \dots + r^{n - 1}) (1 - r)

Distribute $(1 - r)$

S_{n} (1 - r) = a_{1} (1 (1 - r) + r (1 - r) + r^{2} (1 - r) + \dots + r^{n - 1} (1 - r))

Multiply

S_{n} (1 - r) = a_{1} (1 - r + r - r^{2} + r^{2} - r^{3} + \dots + r^{n - 1} - r^{n})

Notice that the like terms are ordered in minus-plus pairs. This means that after simplifying, they will cancel out and only the first and last terms will remain.

S_{n} (1 - r) = a_{1} (1 - r + r - r^{2} + r^{2} - r^{3} + \dots + r^{n - 2} - r^{n - 1} + r^{n - 1} - r^{n})

Add and subtract terms

S_{n} (1 - r) = a_{1} (1 - r^{n})

$LHS / (1 - r) = RHS / (1 - r)$

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

The formula for the sum of a finite geometric series has been derived.

$S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}$

Proof

Infinite Sum

To derive the formula for the sum of an infinite geometric series with

- 1 < r < 1,

the standard formula for the finite geometric series will be considered.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

Since

r

is a number between

- 1

and

1,

the value of

r^{n}

becomes very small as the value of

n

increases. In other words, it gets closer to

0

as

n

approaches infinity.

r^{n} 1 n \to \infty 0

Therefore,

r^{n} = 0

can be substituted into the standard formula and the resulting equation simplified.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

$r^{n} = 0$ , $n = \infty$

S_{\infty} = \frac{a _{1} ( 1 - 0 )}{1 - r}

Subtract term

S_{\infty} = \frac{a _{1} ( 1 )}{1 - r}

Identity Property of Multiplication

S_{\infty} = \frac{a _{1}}{1 - r}

The formula for the sum of an infinite geometric series with

- 1 < r < 1

has been derived.

$S_{\infty} = \frac{a _{1}}{1 - r}$

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