Sum to Infinity of a Geometric Series

**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.

Re: Sum to Infinity of a Geometric Series

Quote:

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: $\displaystyle 1-\dfrac{1}{1-x} = \dfrac{1-x-1}{1-x} = \dfrac{-x}{1-x}$

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

Re: Sum to Infinity of a Geometric Series

Thanks for clearing that up.