Here's a Video going over finding the Taylor series for
$\displaystyle\frac{1}{(1-x)^2}$ which is similar but not quite what I'm looking for.
I know how to find the Taylor/Maclaurin series of
$\displaystyle\frac{1}{(1-x)^2}$ at $c=0$; but I cannot figure out how to manipulate the Taylor series $x^n$ to fit the function of $$\frac{1}{(x-1)^2} \,\,\,\,at \,\,\,\,c=0$$
I tried the series $n(-x)^{n-1}$ but this isn't quite right.
I can't find anything online about this specific problem as only examples like the prior video pop up.
Best Answer
Use that the series $\sum_{n=0}^\infty x^n$ converges absolutely in $B_1(0)$ and use the Cauchy product to obtain \begin{align*} \frac{1}{(1-x)^2} &= \left( \frac{1}{1-x} \right)^2 \\ &= \left( \sum_{n=0}^\infty x^n \right)^2 \\ &= \sum_{n=0}^\infty \sum_{k=0}^n x^n \\ &= \sum_{n=0}^\infty (n+1) x^n. \end{align*}