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$S_{n}=1−ra_{1}(1−r_{n}) ,$ $r =1$

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.

$S_{∞}=1−ra_{1} ,$ $-1<r<1$

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 $∣r∣≥1$ — then the sum *diverges.* In such cases, there is no sum for the infinite geometric series.

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.
Then, both sides of the equation will be multiplied by $(1−r)$ and the resulting equation simplified.
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.
The formula for the sum of a finite geometric series has been derived.

$a_{1},a_{1}r,a_{1}r_{2},…,a_{1}r_{n−1} $

All the terms will be added to express the series and $a_{1}$ will be factored out.
$S_{n}=a_{1}+a_{1}r+a_{1}r_{2}+⋯+a_{1}r_{n−1}$

FactorOut

Factor out $a_{1}$

$S_{n}=a_{1}(1+r+r_{2}+⋯+r_{n−1})$

$S_{n}=a_{1}(1+r+r_{2}+⋯+r_{n−1})$

MultEqn

$LHS⋅(1−r)=RHS⋅(1−r)$

$S_{n}(1−r)=a_{1}(1+r+r_{2}+⋯+r_{n−1})(1−r)$

Distr

Distribute $(1−r)$

$S_{n}(1−r)=a_{1}(1(1−r)+r(1−r)+r_{2}(1−r)+⋯+r_{n−1}(1−r))$

Multiply

Multiply

$S_{n}(1−r)=a_{1}(1−r+r−r_{2}+r_{2}−r_{3}+⋯+r_{n−1}−r_{n})$

$S_{n}(1−r)=a_{1}(1−r+r−r_{2}+r_{2}−r_{3}+⋯+r_{n−2}−r_{n−1}+r_{n−1}−r_{n})$

AddSubTerms

Add and subtract terms

$S_{n}(1−r)=a_{1}(1−r_{n})$

DivEqn

$LHS/(1−r)=RHS/(1−r)$

$S_{n}=1−ra_{1}(1−r_{n}) $

$S_{n}=1−ra_{1}(1−r_{n}) $

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.
*very* small as the value of $n$ increases. In other words, it gets closer to $0$ as $n$ approaches infinity.
The formula for the sum of an infinite geometric series with $-1<r<1$ has been derived.

$S_{n}=1−ra_{1}(1−r_{n}) $

Since $r$ is a number between $-1$ and $1,$ the value of $r_{n}$ becomes $r_{n}1n→∞ 0 $

Therefore, $r_{n}=0$ can be substituted into the standard formula and the resulting equation simplified. $S_{n}=1−ra_{1}(1−r_{n}) $

SubstituteII

$r_{n}=0$, $n=∞$

$S_{∞}=1−ra_{1}(1−0) $

SubTerm

Subtract term

$S_{∞}=1−ra_{1}(1) $

IdPropMult

Identity Property of Multiplication

$S_{∞}=1−ra_{1} $

$S_{∞}=1−ra_{1} $