Power Series Representation of $\frac{1}{(1+3x)^2}$

Question

Question: Find a power series representation of the function:

$$ f(x) = \frac{1}{(1+3x)^2}$$ through the use of an anti-derivative.

My Approach:

1. Take the integral of f(x), Result ==> $\frac{-1}{3}\frac{1}{(1+3x)}$

2. Take the summation of f(x), Result ==> $\frac{-1}{3} \sum_{n=0}^{\infty}(-3)^nx^n$

3. Take the derivative of the summation of f(x), Result ==> $\sum_{n=0}^{\infty}(-3)^{n-1} nx^{n-1}$

End Result ==> Error, Wrong Answer.

Can someone please explain why this result is wrong in addition to pointing out the fallacies in my logic and giving the "proper" result.

Answer 1

So, a primitive of $f$ is $F(x) = -\frac 13 \frac{1}{1+3x} = -\frac 13 \sum_{n=0}^{\infty}(-1)^n 3^n x^{n}$. This power series is absolutely convergent for $x \in ]-\frac 13, \frac 13[$. In this interval you can derive this series term by term, obtaining a convergent series that coincide with $f(x)$ for $x \in ]-\frac 13, \frac 13[$.

More precisely,

$$ f(x)=F'(x)=\left(-\frac{1}{3} \sum_{n=0}^{\infty}(-1)^n 3^n x^{n}\right)' = \sum_{n=1}^{\infty} (-1)^{n+1} 3^{n-1} n x^{n-1}=\sum_{n=0}^{\infty}(-1)^n (n+1) 3^n x^n. $$

I stress the fact that this is only valid when $x \in ]-\frac 13, \frac 13[$.