Answer
\begin{aligned}
T_{n}(x) =\frac{1}{3}-\frac{1}{9}(x-4)+\frac{1}{27}(x-4)^{2}-\frac{1}{81}(x-4)^{3}+\cdots+\frac{(-1)^n}{3^{n+1}}(x-4)^{n}
\end{aligned}
Work Step by Step
Since
\begin{aligned}
f(x) &=\frac{1}{x-1}, & f(4) &=\frac{1}{3} \\
f^{\prime}(x) &=-\frac{1}{(x-1)^{2}}, & f^{\prime}(4) &=-\frac{1}{9} \\
f^{\prime \prime}(x) &=\frac{2}{(x-1)^{2}}, & f^{\prime \prime \prime}(4) &=-\frac{2}{27} \\
\vdots & & & \vdots \\
f^{(n)}(x) &=\frac{(-1)^{n} \cdot n !}{(x-1)^{n+1}} & & f^{(n)}(4)=\frac{(-1)^{n} \cdot n !}{3^{n+1}}
\end{aligned}
Then
\begin{aligned}
T_{n}(x) &=f(4)+\frac{f^{\prime}(4)}{1 !}(x-4)+\frac{f^{\prime \prime}(4)}{2 !}(x-4)^{2}+\cdots+(x-4)^{n} \frac{f^{(n)}(4)}{n !} \\
&=\frac{1}{3}-\frac{1}{9}(x-4)+\frac{1}{27}(x-4)^{2}-\frac{1}{81}(x-4)^{3}+\cdots+\frac{(-1)^n}{3^{n+1}}(x-4)^{n}
\end{aligned}
