Using differentiation to find the power series of a fairly tricky function!! (b) Use part (a) to find a power series for: $$f(x) = \frac{1}{(8+x)^3}$$
What is its radius of convergence?

From Using differentation to find a power series expression for a function We know that
$$\frac{1}{(8+x)^2}=\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n}  \frac{1}{8^{n+2}}$$
We do some reverse engineering to see that
$$\frac{1}{(8+x)^3}=\frac{-1}{2} \frac{d}{dx}\frac{1}{(8+x)^2}$$
Plugging in our answer from part (a):
$$= \frac{-1}{2} \frac{d}{dx}\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n}  \frac{1}{8^{n+2}}$$
Factoring out $\frac{1}{8^2}$ from our sum:
$$= \frac{-1}{2\cdot8^2} \frac{d}{dx}\sum_{n=0}^\infty (-1)^{n} (n+1) \left(\frac{x}{8}\right)^n$$
Taking the derivative yields:
$$= \frac{-1}{2\cdot8^2} \sum_{n=1}^\infty (-1)^{n} (n+1)n \left(\frac{x}{8}\right)^{n-1}\left(\frac{1}{8}\right)$$
The term when $n=0$ is $0$ when we take the derivative, so now our summations starts at $n=1$.
But we want it to start at $n=0$, so we reindex by adding $1$ to everywhere $n$ is:
$$= \frac{-1}{2\cdot8^2} \sum_{n=1}^\infty (-1)^{n+1} (n+2)(n+1) \left(\frac{x}{8}\right)^{n}\left(\frac{1}{8}\right)$$
Now let's pull that $\frac{-1}{8^2}$ back into the sum and combine all the $\frac{1}{8}$ terms:
$$= \frac{1}{2} \sum_{n=1}^\infty (-1)^{n+2} (n+2)(n+1) x^n \left(\frac{1}{8}\right)^{n+3}$$
Notice that the negative sign we brought in made the is what made the $(-1)^{n+1}$ become $(-1)^{n+2}$. Doing a ratio test will show that the radius of convergence is again $8$.

Looks good to me.
Maybe it's "nicer" to write it as $$f(x)=\sum_{n=0}^{\infty}(-1)^n(n+1)(n+2)\frac{x^n}{2^{3n+10}}$$
