An almost-geometric series, 1/4+2/4^2+3/4^3+4/4^4+…=? , a GRE math subject question - Videos

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• Multiply S by 4 and write that below S. Slide the terms ahead by one place, and subtract terms.
  3S/4 = 1/3
  Voila!

• u could call 1/4=x and then u have the sum of (1/4)*n*(x^n) (|x|<1)which is the derivative of the sum of (1/4)*x^(n+1) =(1/4)*x*x^n=(1/4)*x/(1-x) and the derrivative of that is (1/4)/(1-x)^2 so if we plug back in x=1/4 we get that the sum=4/9, but it's only because I like calculus :)

• I will make a video tomorrow on how to create a general solution to the series n/m^n (hopefully).

• The Math Trinity.

• Beautiful solution! I got the same answer by a slightly different method.

  1/4 + 2/4^2 +3/4^3 + 4/4^4...= S
  1 + 2/4 + 3/4^2 + 4/4^3 + 5/4^4 = 4S (Multiplying by 4 takes away a power and i then shifted the series to the left)

  Subtract series

  -1 + (1-2)/4 + (2-3)/4^2 + (3-4)/4^3 = -3S
  1 + (2-1)/4 + (3-2)/4^2 + (4-3)/4^3 = 3S
  1/(1-1/4) = 3S
  4/3 = 3S
  4/9 = S

• Great work!

• Short and easy to understand quickly, awesome!

• Inspiring to do more math! Can't wait for university to continue in October. :-)

• What is a Z series? Anyway nice approach

• you can also differentiate the power series and get the answer