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

1 Answer
Apr 29, 2018

#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#.