Question

How do you find the power series representation for the function `f(x)=1/((1+x)^2)` ?

Answer 1

By Binomial Series,
`1/{(1+x)^2}=(1+x)^{-2}=sum_{n=0}^{infty}(-1)^n(n+1)x^n`

Let us review the binomial series.
`(1+x)^{alpha}=sum_{n=0}^{infty}C(alpha,n)x^n`,
where `C(alpha,n)` is a binomial coefficient defined by
`C(alpha,n)={alpha(alpha-1)(alpha-2)cdot cdots cdot(alpha-n+1)}/{n!}`

Let us first the binomial coefficients for `f(x)=(1+x)^{-2}`.
Since `alpha=-2`, its binomial coefficient looks like
`C(-2,n)={-2(-2-1)(-2-2)cdot cdots cdot[-2-(n-1)]}/{n!}`
`={-2(-3)(-4)cdots[-(n+1)]}/{n!}`
by factoring out all `-`'s in the numerator,
`={(-1)^n[2cdot3cdot4cdot cdots cdot(n+1)]}/{n!}={(-1)^n(n+1)!}/{n!}`
by dividing the numerator and the denominator by `n!`,
`=(-1)^n(n+1)`

Hence, we have the binomial series
`1/{(1+x)^2}=sum_{n=0}^{infty}(-1)^n(n+1)x^n`.

Answer 2

`1/(1+x)^2=sum_(n=0)^oo(-1)^(n+1)nx^(n-1)` for `absx<1`

Explanation:

We have the standard power series:

`1/(1-color(blue)x)=sum_(n=0)^oocolor(blue)x^n`

From this we write the power series for `1/(1+x)`:

`1/(1+x)=1/(1-color(blue)((-x)))=sum_(n=0)^oo(color(blue)(-x))^n=sum_(n=0)^oo(-1)^nx^n`

Note that `1/(x+1)=(x+1)^-1`, so `d/dx(1/(1+x))=-(x+1)^-2`:

`d/dx(1/(1+x))=(-1)/(x+1)^2=d/dxsum_(n=0)^oo(-1)^nx^n`

Which can be rewritten as:

`(-1)/(1+x)^2=sum_(n=0)^oo(-1)^nd/dxx^n=sum_(n=0)^oo(-1)^n(nx^(n-1))`

Reversing the signs by multiplying both sides by `-1`:

`1/(1+x)^2=-sum_(n=0)^oo(-1)^n(nx^(n-1))=sum_(n=0)^oo(-1)^(n+1)nx^(n-1)`

Which is convergent for `absx<1`.