Problem source

The function $\mathrm{f}$ is defined, for $x\neq1$ and $x\neq2$ by
\[
\mathrm{f}(x)=\frac{1}{\left(x-1\right)\left(x-2\right)}
\]
Show that for $\left|x\right|<1$
\[
\mathrm{f}(x)=\sum_{n=0}^{\infty}x^{n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}
\]
and that for $1<\left|x\right|<2$ 
\[
\mathrm{f}(x)=-\sum_{n=1}^{\infty}x^{-n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}
\]
Find an expression for $\mbox{f}(x)$ which is valid for $\left|x\right|>2$.

Solution source

\begin{align*}
&& \f(x) &= \frac1{(x-1)(x-2)} \\
&&&=\frac{1}{x-2} -\frac{1}{x-1} \\
\end{align*}

Therefore, for $|x| < 1$

\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\
&&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2
\end{align*}
where both geometric series converge since $|x| < 1$ and $|\frac{x}{2}| < 1$

When $1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1$, we must have:

\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&&=  -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\
&&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\
&&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\
\end{align*}

Finally, when $|x| > 2$, ie $|\frac{2}{x}| < 1$ we have
\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\
&&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\
&&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\
\end{align*}