<s>[INST] Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Express your answer in sigma notation, simplified as much as possible.
This is a practice question that I am having trouble with. I know how to get it about $x=0$, but it is the x=1 part that confuses me.
 [/INST] $f(x) 
= \dfrac{1}{(x − 2)^2}
$
Let $y = x-1$,
or $x= y+1$.
We want to find the expansion
about $y = 0$.
$\begin{array}\\
g(y)
&=f(y+1)\\
&= \dfrac{1}{(y-1)^2}\\
&= \dfrac{1}{(1-y)^2}\\
&=\sum_{n=0}^{\infty} (n+1)y^n\\
&=\sum_{n=0}^{\infty} (n+1)(x-1)^n\\
&\text{and we could stop here, but, for the fun of it,...}\\
&=\sum_{n=0}^{\infty} (n+1)\sum_{k=0}^n \binom{n}{k}x^k(-1)^{n-k}\\
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty} (n+1) \binom{n}{k}x^k(-1)^{n-k}\\
&=\sum_{k=0}^{\infty}x^k(-1)^k\sum_{n=k}^{\infty} (n+1) \binom{n}{k}(-1)^{n}\\
&=\sum_{k=0}^{\infty}x^k(-1)^k\sum_{n=k}^{\infty} \frac{(n+1)!}{k!(n-k)!}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=k}^{\infty} \frac{(n+1)!}{(k+1)!(n-k)!}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=k}^{\infty} \binom{n+1}{k+1}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=0}^{\infty} \binom{n+k+1}{k+1}(-1)^{n}\\
&\text{and this is clearly divergent, so please disregard the last 7 lines}\\
\end{array}
$
</s>