I have the following power series:
$f(T) = \sum_{n\geq 0}(n+1)T^n \in \mathbb{C} [T]$
and I have to compute the multiplicate inverse.
I've tryed to use the formula to calculate the coefficients of the inverse one by one in order to find if there were any rule but it wasnt usefull.
$b_0= a_0^{-1}$
$b_n = -a_0^{-1}\sum_{i=1}^{n}a_ib_{n-i}$
then $b_0= 1, b_1 = -2, b_2 = 1$ and $b_n=0 \space \forall n \geq 3$
Thank you in advance.
I rewrite the prove of Conrad.
Let $g(T)=\sum_{n\geq1}T^n$ then $g'(T)=f(T)$
It is known that the multiplicative inverse of $\sum_{n\geq0}T^n$ is $(1-T)$ then
$(g(T)+1)(1-T)=1;g(T)+1=\frac{1}{1-T}$ and taking the derivative
$g'(T)=f(T)=\frac{1}{(1-T)^2}$
then the multiplicative inverse of $f$ is $(1-T)^2$