Answer:
x = 0
Step-by-step explanation:
Given:
[tex]\displaystyle \large{S = 1+x+x^2+x^3+... \to (1)}\\\displaystyle \large{S=1-x+x^2-x^3+... \to (2)}[/tex]
Convergent Definition / Infinite Geometric Series:
[tex]\displaystyle \large{S=\dfrac{a_1}{1-r} \ \ \tt{for \ |r| < 1}}[/tex]
- S = sum
- [tex]\displaystyle \large{a_1}[/tex] = first term
- r = common ratio
From (1):-
Our common ratio is x and first term is 1:
[tex]\displaystyle \large{S=\dfrac{1}{1-x}}[/tex]
From (2):-
Our common ratio is -x and first term is 1:
[tex]\displaystyle \large{S=\dfrac{1}{1+x}}[/tex]
To find:
- x-value(s) that make both series equal to each other.
So we solve the equation between two series:
[tex]\displaystyle \large{\dfrac{1}{1-x}=\dfrac{1}{1+x}}[/tex]
|x| < 1 since it’s convergent so |x| cannot be greater than 1 or less than -1.
Solve the equation:
[tex]\displaystyle \large{\dfrac{1}{1-x}(1+x)(1-x) = \dfrac{1}{1+x}(1-x)(1+x)}\\\displaystyle \large{1(1+x)=1(1-x)}\\\displaystyle \large{1+x=1-x}\\\displaystyle \large{2x=0}\\\displaystyle \large{x=0}[/tex]
Therefore, the only possible x-value for both convergent sum to be equal is x = 0
