Question

How do you find the maclaurin series expansion of `f(x) = x/(1-x)`?

Answer 1

If `|x|<1`, the expansion is `x+x^2+x^3+...=sum_{i=1}^infty x^i`

Explanation:

This is already a known expansion, since `sum_{i=1}^infty x^i=x/(1-x)`.

To see why this is true, you need to compute the derivatives and evaluate them in zero, and then use the expression

`f(x)= f(0) + f'(0)* x + (f''(0)) /(2!) * x^2 + (f'''(0)) /(3!)x^3+...`

and verify that every coefficient `f^{(n)}(0)/(n!)` equals one.

If you're only interested in what you had to do to find out the expansion, that's it. If you want a proof of the last mentioned fact, don't hesitate to ask