= Solution
Set $a=\sqrt{m\omega/(2\hbar)}x+\sqrt{\hbar/(2m\omega)}\,d/dx$ and let $a^\dagger$ have the opposite derivative sign. On the bound-state domain, $[a,a^\dagger]=1$ and $H=\hbar\omega(a^\dagger a+1/2)$. Thus every <energy eigenvalue> has $E/\hbar\omega-1/2=\nu\geq0$.
The <lowering operator> maps an eigenstate of $a^\dagger a$ of <eigenvalue> $\nu$ to one of <eigenvalue> $\nu-1$, with squared <norm> multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a\psi_0=0$ gives a Gaussian <ground state>, and repeated raising gives a nonzero <square-integrable> state for every integer $n\geq0$. Therefore
$$
\boxed{E_n=\hbar\omega(n+1/2),\qquad n=0,1,2,\ldots.}
$$
These are also the Gaussian-times-Hermite-<polynomial> solutions obtained by the suggested series method.
Put $\alpha=m\omega/\hbar$. Normalization gives $\psi_0=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}$, so
$$
\langle|x|\rangle=2\sqrt{\frac\alpha\pi}\int_0^\infty xe^{-\alpha x^2}dx
=\boxed{\sqrt{\frac\hbar{\pi m\omega}}}.
$$
The <expectation> of $x$ itself is zero by <parity>; that cancellation does not apply to its absolute value.
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