Answer
$T (x)=\Sigma_{n = 0}^{\infty} (-1)^n(n+1)\dfrac{(x-4)^n}{4^{n+2}}$; converges for all $x$ with convergence interval $(0,8)$.
Work Step by Step
The Taylor`s Series can be represented as: : $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$
If we consider $c=0$, then Taylor`s Series, $T(x)$ is also called as Maclaurin Series and the Maclaurin series for $f(x)$ is as given below:
$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+..............$
Here, we have
$f(x)=\dfrac{1}{x^2} \implies f(4)=\dfrac{1}{16}\\ f'(x)=-2x^{-3} \implies f'(4)=\dfrac{-1}{32}\\f''(x)=-6x^{-4} \implies f''(4)=\dfrac{3}{128}\\f'''(x)=-24x^{-5} \implies f'''(4)=\dfrac{-3}{128}\\f''''(0)=120x^{-6} \implies f^{iv}(4)=\dfrac{15}{512}$
Thus, we have the Taylor`s Series: $T (x)=\dfrac{1}{16}-\dfrac{1}{32}(x-4)+\dfrac{3}{128}\dfrac{(x-4)^2}{2!}+..............\\=\Sigma_{n = 0}^{\infty} (-1)^n(n+1)\dfrac{(x-4)^n}{4^{n+2}}$; converges for all $x$ with convergence interval $(0,8)$.
