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

##### 1 Answer

#### Explanation:

We start with the well known series:

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

Replace

#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)^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

We also multiply by a factor of

#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