<s>[INST] Using differentation to find a power series expression for a function (a) use differentiation to find a power series representation for: $$f(x) = \frac{1}{(8+x)^2}$$
What is it's radius of convergence?

Part (a):
We start off with some tricky reverse engineering of derivatives:
$\frac{1}{(8+x)^2} = \frac{d}{dx}(\frac{-1}{x+8}) = \frac{d}{dx} ((\frac{-1}{8})(\frac{1}{\frac{x}{8}+1})) = \frac{d}{dx} ((\frac{-1}{8})(\frac{1}{1-\frac{-x}{8}}))$
Great! Now we can turn $(\frac{1}{1-\frac{-x}{8}}))$ into a power series to get :
$=\frac{d}{dx} ((\frac{-1}{8})\sum_{n=0}^\infty (-1)^n (\frac{x}{8})^n)$
Now we take the derivative....
$=(\frac{-1}{8})\sum_{n=1}^\infty (-1)^n  n (\frac{x}{8})^{n-1}(\frac{1}{8})$
We are starting at $n=1$ now because the term when $n=0$ becomes $0$ when we take the derivative.
Combining all the $\frac{1}{8}$ we get:
$=\sum_{n=1}^\infty (-1)^{n+1} n x^{n-1}  \frac{1}{8^{n+1}}$
notice that the alternating term went up to ${n+1}$ because of the negative in the term $\frac{-1}{8}$ out in front.
But the answer wants us to start this series at $n=0$, so we must reindex. So, if we change it so $n$ starts at $0$ rather than starting at $1$, that would means we have to add $1$ to every $n$ in our expression:
$=\sum_{n=0}^\infty (-1)^{n+2} (n+1) x^{n}  \frac{1}{8^{n+2}}$
But $(-1)^{n+2} = (-1)^n$ for all $n$, so this equals:
$=\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n}  \frac{1}{8^{n+2}}$
Applying the ratio test you will see that this converges when $|x| < 8$, so $R = 8$
Note: Before when we were using the alternating series test, it didn't matter the form of the alternating term... It could have been $(-1)^n$ or $(-1)^{n-1}) for example.. Now it matters!!
 [/INST] You are correct.
Alternatively since $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
for $|x|<1$ then $\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$. So we have $$\frac{1}{(8+x)^2}=\frac{1}{8^2}\frac{1}{(1-(-\frac{x}{8}))^2}$$
$$=\frac{1}{8^2}\sum_{n=1}^{\infty}n\big(-\frac{x}{8}\big)^{n-1}=\sum_{n=1}^{\infty}(-1)^{n-1}nx^{n-1}\frac{1}{8^{n+1}}$$
$$=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n\frac{1}{8^{n+2}}$$
which converges for $|\frac{x}{8}|<1$ or $|x|<8$.
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