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

1 Answer
Apr 29, 2018

1/(1+x)^2=sum_(k=1)^oo(-1)^(k+1)kx^(k-1)1(1+x)2=k=1(1)k+1kxk1 with R=1R=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.