Question

How to find the Maclaurin series and the radius of convergence for `f(x)=1/(1+x)^2`?

Answer 1

`1/(1+x)^2=sum_(k=1)^oo(-1)^(k+1)kx^(k-1)` with `R=1`

Explanation:

We start with the well known series:

`1/(1-x)=sum_(k=0)^oox^k=1+x+x^2+x^3+...`

Replace `x` with `(-x)`:

`1/(1+x)=1/(1-(-x))=sum_(k=0)^oo(-x)^k=sum_(k=0)^oo(-1)^kx^k`

Note that the derivative of `1/(1+x)=(1+x)^-1` is `-(1+x)^-2=(-1)/(1+x)^2`. We should then differentiate the series:

`(-1)/(1+x)^2=d/dxsum_(k=0)^oo(-1)^kx^k=sum_(n=0)^oo(-1)^kd/dxx^k=sum_(k=0)^oo(-1)^k(kx^(k-1))`

We can move the index of the sum to start at `k=1`, since the term `k=0` results in an addend of just `0`.

We also multiply by a factor of `-1` to find the desired result:

`1/(1+x)^2=-sum_(k=1)^oo(-1)^kkx^(k-1)=sum_(k=1)^oo(-1)^(k+1)kx^(k-1)`

`=1-2x+3x^2-4x^3+...`

Which has radius of convergence `R=1`, since the original substitution of `-x` for `x` would not affect the radius, and neither would the differentiation or multiplication by a factor of `-1`.