Sum to Infinity of a Geometric Series

**GrigOrig99** · August 7th 2011, 01:04 PM

Q. Find in terms of x, in its simplest form, the sum of the infinite geometric series

x^2 + x^2/ (1 - x) + x^2/ (1 - x)^2 + x^2/ (1 - x)^3 +…

Eq.: S¥ = a/ 1 - r

Ans.: From text-book, x(x - 1)

Attempt: a = x^2, r = 1/ (1 - x)

x^2/ (1 - (1/ (1 - x)))
x^2(-1 + x) …. inverting 1/ (1 - x) and changing +/ - via the 1 - from 1 - 1/(1 - x)
x^3 - x^2
x^2(x - 1)

This is as close to the answer as I can get. I can’t quite work it out and was hoping someone could help. Thank you.

**e^(i*pi)** · August 7th 2011, 01:08 PM

Originally Posted by GrigOrig99

Q. Find in terms of x, in its simplest form, the sum of the infinite geometric series

x^2 + x^2/ (1 - x) + x^2/ (1 - x)^2 + x^2/ (1 - x)^3 +…

Eq.: S¥ = a/ 1 - r

Ans.: From text-book, x(x - 1)

Attempt: a = x^2, r = 1/ (1 - x)

x^2/ (1 - (1/ (1 - x)))
x^2(-1 + x) …. inverting 1/ (1 - x) and changing +/ - via the 1 - from 1 - 1/(1 - x)
x^3 - x^2
x^2(x - 1)

This is as close to the answer as I can get. I can’t quite work it out and was hoping someone could help. Thank you.

Your a and r values are fine.

If we look at the denominator (1-r) and simplify it: $1-\dfrac{1}{1-x} = \dfrac{1-x-1}{1-x} = \dfrac{-x}{1-x}$

This makes our sum $S_{\infty} = \dfrac{x^2}{\left(\dfrac{-x}{1-x}\right)} = x^2 \cdot \dfrac{1-x}{-x} = -x(1-x) = x(x-1)$

**GrigOrig99** · August 7th 2011, 02:10 PM

Thanks for clearing that up.

Thread: Sum to Infinity of a Geometric Series

LinkBack

Thread Tools

Display