<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] Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad\qquad \alpha\in\mathbb{C}, |x|<1
\end{align*}

We obtain
  \begin{align*}
\frac{1+x}{(1-x)^2}&=(1+x)\sum_{n=0}^{\infty}\binom{-2}{n}(-x)^n\tag{1}\\
&=(1+x)\sum_{n=0}^{\infty}\binom{n+1}{n}x^n\\
&=(1+x)\sum_{n=0}^{\infty}(n+1)x^n\\
&=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=0}^{\infty}(n+1)x^{n+1}\tag{2}\\
&=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=1}^{\infty}nx^{n}\\
&=\sum_{n=0}^{\infty}(2n+1)x^n
\end{align*}

Comment:


*

*In (1) we use the identity
$\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$

*In (2) we shift the index of the right sum by one
</s>