<s>[INST] Power series of $\frac{1+x}{(1-x)^2}$ This question is continuing from the previous question here:
Power Series representation of $\frac{1+x}{(1-x)^2}$
I am trying to calculate the power series representation of the equation:
$$
\begin{align}
f(x) = \frac{1+x}{(1-x)^2}
\end{align}
$$
My workout is as follow:
$$
\begin{align}
\frac{1+x}{(1-x)^2} = \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2}
\end{align}
$$
For $\frac{1}{(1-x)^2}$:
\begin{align}
 \frac{1}{(1-x)^2} &= \frac{d}{dx} \frac{1}{1-x}\\
&= \frac{d}{dx} \sum_{n=0}^{\infty} x^n \\
&= \sum_{n=1}^{\infty} nx^{n-1} \\
&= \sum_{n=0}^{\infty} (n+1)x^n
\end{align}
For$\frac{x}{(1-x)^2}$:
$$
\begin{align}
x \frac{1}{(1-x)^2} &= x \sum_{n=0}^{\infty}(n+1)x^n \\
&= \sum_{n=0}^{\infty} (n+1) x^{n+1}
\end{align}
$$
Therefore, $$
\begin{align}\frac{1+x}{(1-x)^2}
= \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1}),\end{align}
$$
where range of convergence is $x\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\infty)(0) = 0$.
However, the model answer is $\sum_{n=0}^{\infty} (2n+1) x^n$, where range of convergence is $x\in (-1,1)$.  
I do not understand what is wrong with my calculation. Any advice will be appreciated!
 [/INST] You did not end up with power series yet:
\begin{align*}
\sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1}  &=\sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=1}^{\infty} n x^{n}\\
&=\sum_{n=1}^{\infty} (2n+1) x^{n} + 1\\
&=\sum_{n=0}^{\infty} (2n+1) x^{n}\\
\end{align*}
From here you can derive correct radius of convergence.
</s>