Answer:
x = 0
Explanation:
Given:
![\displaystyle \large{S = 1+x+x^2+x^3+... \to (1)}\\\displaystyle \large{S=1-x+x^2-x^3+... \to (2)}](https://img.qammunity.org/2022/formulas/mathematics/college/5fb8jt6jlsdfa7kqde4yg3ge9szdcwf176.png)
Convergent Definition / Infinite Geometric Series:
![\displaystyle \large{S=(a_1)/(1-r) \ \ \ttr}](https://img.qammunity.org/2022/formulas/mathematics/college/o4mc2rdhd281fpv03q4yycnf8m56z6bm3p.png)
- S = sum
= first term- r = common ratio
From (1):-
Our common ratio is x and first term is 1:
![\displaystyle \large{S=(1)/(1-x)}](https://img.qammunity.org/2022/formulas/mathematics/college/ftswhcn9iye20ku9pgj2rc26ujpu2wr0a7.png)
From (2):-
Our common ratio is -x and first term is 1:
![\displaystyle \large{S=(1)/(1+x)}](https://img.qammunity.org/2022/formulas/mathematics/college/salv2sc10un58eli1w7qf3jgy8akm45uvr.png)
To find:
- x-value(s) that make both series equal to each other.
So we solve the equation between two series:
![\displaystyle \large{(1)/(1-x)=(1)/(1+x)}](https://img.qammunity.org/2022/formulas/mathematics/college/muxv8s6tn8zci6sj71ge579gv401hmhvvl.png)
|x| < 1 since it’s convergent so |x| cannot be greater than 1 or less than -1.
Solve the equation:
![\displaystyle \large{(1)/(1-x)(1+x)(1-x) = (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}](https://img.qammunity.org/2022/formulas/mathematics/college/j85z2yf0svod9mm5xdxb7e59qyq7dx53ww.png)
Therefore, the only possible x-value for both convergent sum to be equal is x = 0