power series

${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n=c_0+c_1(x-a)+c_2(x-a{)}^2+\dots$

1. series converges for all $x=a$, diverges for all $x\ne a$
2. series converges for all real numbers $x$
3. exists a real number $R>0$ s.t. converges if $|x-a|<R$ and diverges if $|x-a|>R$. at $|x-a|=R$, may converge or diverge

if there exists a real number $d\ne 0$ such that ${\sum }_{n=0}^{\infty }{c}_n{d}^n$ converges, then the series ${\sum }_{n=0}^{\infty }{c}_n{x}^n$ converges absolutely for all $x$ such that $|x|<|d|$

since ${\sum }_{n=0}^{\infty }{c}_n{d}^n$ converges, the $n$th term $c_n{d}^n\to 0$ as $n\to \infty$
therefore, exists an integer $N$ such that $|c_n{d}^n|\le 1$ for all $n\ge N$

turning $d$ into $x$, $|c_n{x}^n|=|c_n{d}^n||\frac{x}{d}{|}^n$

and since limiting $n$ to having $|c_n{d}^n|\le 1$
$|c_n{x}^n|\le |c_n{d}^n|\cdot |\frac{x}{d}{|}^n\le 1\cdot |\frac{x}{d}{|}^n$ holds
so $|c_n{x}^n|\le |\frac{x}{d}{|}^n$

the series ${\sum }_{n=N}^{\infty }|c_n{x}^n|\le {\sum }_{n=N}^{\infty }|\frac{x}{d}{|}^n$ converges if $|\frac{x}{d}|\le 1$, which it is
so ${\sum }_{n=N}^{\infty }|c_n{x}^n|$ must converge based on being squeezed by $\le$

$f(x)=\frac{x^3}{2-x}$ into power series

$\frac{x^3}{2-x}=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots$

moving $2-x$ to right
multiplying $2-x$ by each term gives $2c_n{x}^n$ and $-c_n{x}^{n+1}$
rearranging, $\frac{x^3}{2-x}=2c_0+(2c_1-c_0)x+(2c_2-c_1)x^2+(2c_3-c_2)x^3+\dots$

removing terms, so $c_{\mathrm{0..2}}=0$
but need $x^3$ term, so $c_3=\frac{1}{2}$ (as $2c_3=1$)
in $x^4$ term, need to get rid of $-c_3$ now, so $c_4=\frac{1}{4}$ (as $2c_4=c_3=\frac{1}{2}$)
in $x^5$ term, need to get rid of $c_4$ and so on…

$c_{\mathrm{0..2}}=0$
$c_{n\ge 3}=\frac{1}{2^{n-2}}$

$\frac{x^3}{2-x}={\sum }_{k=3}^{\infty }\frac{1}{2^{k-2}}{x}^k=4{\sum }_{k=3}^{\infty }(\frac{x}{2}{)}^k$

converges at $|\frac{x}{2}|<1$, $|x|<2$
interval of convergence is $(-2,2)$