<s>[INST] Writing a function in terms of its power series with some tricky reindexing steps Represent the following function as a power series and find it's radius of convergence:
$$\frac{x^2}{(8+x)^3}$$

By Using differentiation to find the power series of a fairly tricky function!! We know that
$\frac{1}{(8+x)^3} = \frac{1}{2} (\sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^n (\frac{1}{8})^{n+3})$
Therefore we have:
$\frac{x^2}{(8+x)^3} = x^2 (\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^n (\frac{1}{8})^{n+3})$
Multiplying the $x^2$ through yields
$= \frac{1}{2} (\sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^{n+2} (\frac{1}{8})^{n+3})$
But we can't have a power series be expressed in terms of $x^{n+2}$. It needs to be in terms of $x^n$. So we subtract a $2$ from everywhere an $n$ appears in our expressoin and make the summation start at $n=2$
$= \frac{1}{2} (\sum_{n=2}^\infty (-1)^{n} (n)(n-1) x^{n} (\frac{1}{8})^{n+1})$
 [/INST] Yes it's correct. Since for $n=0$ and $n=1$ the terms vanish, if you want to start at $n=0$ we can write $$\frac{x^2}{(x+3)^3}= \frac{1}{2} \sum_{n=2}^\infty (-1)^{n} (n)(n-1) x^{n} (\frac{1}{8})^{n+1}=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n(n)(n-1)x^n(\frac{1}{8})^{n+1}$$
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