The webpage title could be: Solving for x in an Infinite Geometric Series

[Niaboc67]

x+x^2+x^3+x^4... = 14

Find x

Could someone please provide an explanation.

Thank you

 

[JonnyG]

1 + x + x^2 +x^3 + ... = 1/(1-x) for |x| < 1

 

[Niaboc67]

Could you explain why is it 1/(x-1) and for the abs x <1. I don't understand the reason for these and geometric series.
Thanks

 

[JonnyG]

1 + x + x^2 + ... + x^(n-1) = (1 - x^n)/(1-x)

This expression is valid for all x not equal to 1. Now let n go to infinity. The right side converges to 1/(1-x) if and only if abs(x) < 1 since x^n will goes to 0 for x in the interval (-1,1) and will diverge for x <= -1 or x > 1

 

[HallsofIvy]

A simpler way, I think: x+ x^2+ x^3+ ...= 14.
Factor out an x: x(1+ x+ x^2+ x^3+ ...)= x(1+ (x+ x^2+ x^3+ ...))= x(1+ 14)= 15x= 14 so x= 14/15.