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Question: How do you find the sum of the infinite geometric series \(1 - x + {x^2} - {x^3} + {x^4} - ...?\)...

How do you find the sum of the infinite geometric series 1−x+x2−x3+x4−...?1 - x + {x^2} - {x^3} + {x^4} - ...?

Explanation

Solution

An infinite geometric series is the series given and, as we know, the common ratio of consecutive terms in a geometric series is fixed. To find the sum of infinite geometric series, firstly, by dividing a term with its progressive term, find the common ratio between terms. And then use the following formula:
S∞=a1−r,  where  S∞,  a  and  r{S_\infty } = \dfrac{a}{{1 - r}},\;{\text{where}}\;{S_\infty },\;a\;{\text{and}}\;r are sum of infinite geometric series, first term of the series and common ratio of the series respectively.

Formula used:
Common ratio of a G.P.: r=un+1unr = \dfrac{{{u_{n + 1}}}}{{{u_n}}}
Infinite sum of G.P.: S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}

Complete step by step solution:
In order to find the sum of the infinite geometric series 1−x+x2−x3+x4−...1 - x + {x^2} - {x^3} + {x^4} - ... we will first find the common ratio of the series as following
r=un+1un,  where  un+1  and  unr = \dfrac{{{u_{n + 1}}}}{{{u_n}}},\;{\text{where}}\;{u_{n + 1}}\;{\text{and}}\;{u_n} are “n+1th” and “nth” term of the geometric series respectively
We will take second and first term, to find the common ratio,
⇒r=−x1=−x\Rightarrow r = \dfrac{{ - x}}{1} = - x
Now, we will use the formula for sum of infinite terms of geometric series which is given as follows
S∞=a1−r,  where  S∞,  a  and  r{S_\infty } = \dfrac{a}{{1 - r}},\;{\text{where}}\;{S_\infty },\;a\;{\text{and}}\;r are sum of infinite geometric series, first term of the series and common ratio of the series respectively
In the series 1−x+x2−x3+x4−...1 - x + {x^2} - {x^3} + {x^4} - ... the first term is a=1a = 1
Putting a=1  and r=−xa = 1\;{\text{and}}\,r = - x in the above formula, we will get
S∞=11−(−x)=11+x{S_\infty } = \dfrac{1}{{1 - ( - x)}} = \dfrac{1}{{1 + x}}
Therefore the required infinite sum of the given series is equals to 11+x\dfrac{1}{{1 + x}}

Note: The infinite sum of the following series will only exist when value of “x” will be less than one and greater than negative one excluding zero, that is x∈(−1,  1)∼0x \in \left( { - 1,\;1} \right)\sim0, otherwise the infinite sum of the given series will not exist and also if x=0x = 0 then series itself will not exist.