Expand the function as Taylor series Expand given function $f$ as Taylor series around $c=3$
$$f(x) = \frac{x-3}{(x-1)^2}+\ln{(2x-4)} $$

and find out open interval at what that series is convergent. What is radius of convergence?

This is what i have so far. We know that $\ln{(1+x)} = \sum_{n=1}^{\infty} \frac{x^n}{n}$ and $\frac{1}{1-x}=\sum_{n=1}^\infty x^n ,|x| \lt 1$

We can rewrite $\ln{(2x-4)}$ as $\ln{(-\frac{1}{4})}+\ln{(-x/2+1)}$ and then expand last expression, but i don't know what to do with $\ln{(-1/4)}$. We can can split given fraction onto partial fractions as $\frac{x-3}{(x-1)^2} = \frac{1}{x-1}-\frac{2}{(x-1)^2}$. First fraction we can expand easily, but i don't know how to expand fraction with binomial as denominator.
We have that for $|z|&lt;1$,
$$\frac{1}{(1-z)^2}=\left(\frac{1}{1-z}\right)'=\left(\sum_{n\geq 0} z^{n}\right)'=\sum_{n\geq 1}n z^{n-1}.$$
Let $t=x-3$, then for $t\in (-1,1]$
$$\frac{x-3}{(x-1)^2}+\ln{(2x-4)}=\frac{t}{(t+2)^2}+\ln(2)+\ln(1+t)\\
=\frac{t}{4(1-(-t/2))^2}+\ln(2)+\ln(1+t)\\
=\frac{t}{4}\sum_{n\geq 1}n (-t/2)^{n-1}+\ln(2)-\sum_{n\geq 1}\frac{(-t)^n}{n}\\
=\ln(2)+\sum_{n\geq 1} t^n\cdot(-1)^{n-1}\left(\frac{n}{2^{n+1}}+\frac{1}{n}\right).
$$
Note that the radius of convergence is $1$, that is the smaller radius between the radius of the series of $\ln(1+t)$, that is $1$,  and the radius of the series of $\frac{1}{(1-(-t/2))^2}$, that is $2$.
$$f(x)=\frac{x-1-2}{(x-1)^2}+\ln2+\ln(x-2)=\ln2+\frac1{x-1}-\frac2{(x-1)^2}+\ln(x-2)$$ and $f(3)=\ln2$.

Then

$$f'(x)=-\frac1{(x-1)^2}+\frac4{(x-1)^3}+\frac1{x-2}.$$

Next
$$f''(x)=\frac2{(x-1)^3}-\frac{12}{(x-1)^4}-\frac1{(x-2)^2}.$$

The general pattern isn't difficult to find out,

$$f^{(k)}(x)=(-1)^k\left(\frac{k!}{(x-1)^{k+1}}-2\frac{(k+1)!}{(x-1)^{k+2}}+\frac{(k-1)!}{(x-2)^k}\right).$$

Finally, for $k&gt;0$,

$$a_k=\frac{f^{(k)}(3)}{k!}=(-1)^k\left(\frac1{2^{k+1}}-2\frac{k+1}{2^{k+2}}+\frac1k\right).$$