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Infinite geometric series

  1. Aug 4, 2015 #1
    x+x^2+x^3+x^4.... = 14

    Find x

    Could someone please provide an explanation.

    Thank you
     
  2. jcsd
  3. Aug 4, 2015 #2
    1 + x + x^2 +x^3 + ... = 1/(1-x) for |x| < 1
     
  4. Aug 4, 2015 #3
    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
     
  5. Aug 4, 2015 #4
    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
     
  6. Aug 4, 2015 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
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