Find the Taylor Series for $f(x)$ centered at a given value $a$ $$f(x) = \frac{6}{x}\,\, \mathrm{at}\,\, a = -4 .$$
Assume that $f$ has a power series expansion. Do not show that $R_n(x) -> 0$
I took the derivatives of f(x):
$$f(x) = 6/x$$
$$f'(x) = -6/x^2$$
$$ f''(x) = 12/x^3 $$
$$ f^{(3)}(x) = -36/x^4 $$
but I am unable to write out the series, please help! Thank you
 A: For a function $f$, the definition of the Taylor Series of $f$ is:
$$\text{Taylor}(f)=\sum_{n \ge 0}\frac{f^{(n)}(a)}{n!}(x-a)^n, $$
where $f^{(n)}$ indicates the $n$th derivative of $f$ with $f^{(0)}=f$.
The problem, then, reduces to the following: Is there an $n$th derivative of $f(x)=\frac{6}{x}$? (More aptly, since $6$ is a constant: Is there an $n$th derivative of $\frac{1}{x}$? If so, what is it?)
Let's do a table of values:
$$
\begin{align}
\frac{d}{dx}\frac{1}{x}&=-\frac{1}{x^2}\\
\frac{d^2}{dx^2}\frac{1}{x}&=2\frac{1}{x^3}\\
\frac{d^3}{dx^3}\frac{1}{x}&=-6\frac{1}{x^4}\\
&\ldots
\end{align}
$$
This pattern seems to suggest $$\frac{d^n}{dx^n}\frac{1}{x}=(-1)^{n}n!\frac{1}{x^{n+1}}.$$
Let's prove this via induction on $n$. The base case trivially true. Assume the induction hypothesis for $n=k$.
$$
\begin{align}
\frac{d^{k+1}}{dx^{k+1}}\frac{1}{x}&=\frac{d}{dx}\frac{d^k}{dx^k}\frac{1}{x}\\
&=\frac{d}{dx}(-1)^kk!\frac{1}{x^{k+1}}\\
&=(-1)^kk!(k+1)(-1)\frac{1}{x^{k+2}}\\
&=(-1)^{k+1}(k+1)!\frac{1}{x^{k+2}}.
\end{align}
$$
Thus, via induction, we see this is correct!
Can you take this from here? Apply the $n$th derivative and the definition of $\text{Taylor}(f)$.

Okay, so you have the following pieces of information: $a=-4$, $f^{(n)}(x)=(-1)^nn!\frac{1}{x^{n+1}}$, and the definition of the Taylor Series. You simply substitute these into the definition as follows:
$$
\text{Taylor}(f)=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{a^{n+1}}}{n!}(x-a)^n=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{(-4)^{n+1}}}{n!}(x-(-4))^n . . .
$$
Try to go from here and simplify this. :)
A: Refer to the definition of Taylor Series. You will have to evaluate your derivatives at $a=-4$.
