Why cant I square the series when doing maclaurin expansion So the task says: Do the Maclaurin expansion of the given term: $$f(x) = \frac{x}{(1-x)^2}$$
what I did was: https://ibb.co/19Ddd8d
The way i tried to solve the task was wrong, I know that i should derive and than multiply with x and that would give me the correct Maclaurin expansion. 
But why couldn't I square both sides and then multiply with x instead of deriving  and than multiplying with x?
 A: Perhaps you should sniff around some texts about generating functions to get more comfortable with these calculations. You can certainly square both sides and multiply by $x$. For example:
$$\frac{1}{1-x} = \sum_{m=0}^{\infty} x^m$$
Squaring both sides yields
\begin{align}
\frac{1}{(1-x)^2} &= \left(\sum_{m=0}^{\infty} x^m\right)^2\\
&= (1+x+x^2+\cdots )(1+x+x^2+\cdots)\\
&= (1+2x+3x^2+4x^3+\cdots)\\
&= \sum_{m=0}^{\infty} (m+1)x^m.
\end{align}
Notice that you'd need to calculate the square of the polynomial. Fortunately it is easy enough in this case.
Finally multiplying by $x$ yields
\begin{align}
\frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} (m+1)x^{m+1}\\
&= \sum_{m=1}^{\infty} mx^{m}
\end{align}
On the other hand, we can take a derivative and be done with it. Taking the derivative of the right hand side is fairly straightforward, as we just take the derivative of every term individually.
\begin{align}
\frac{1}{1-x}\frac{d}{dx} &= \sum_{m=0}^{\infty} x^m\frac{d}{dx}\\
\frac{1}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m-1}\\
\frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m}\\
\frac{x}{(1-x)^2} &= \sum_{m=1}^{\infty} mx^{m}\\
\end{align}
I forget if one should really take caution while taking the derivative of a sum $\sum_{m=0}^{\infty} x^m$ as the first term without adjusting indices becomes $0(\frac{1}{x})$, but perhaps someone with more experience can speak to this. 
