To elaborate on some of the above points, consider differencing a process following a deterministic trend $y_t=a+bt+\epsilon_t$.
$\Delta y_t$ is not invertible, as $\Delta y_t=bt-b(t-1)+\Delta\epsilon_t=b+\Delta\epsilon_t$. This is an $MA(1)$ with a unit root, and hence not invertible. This is because first differencing is the "wrong" detrending scheme for a trend stationary process.
Also, we have that the long-run variance of an $MA(1)$ process can be written as
$$
J=\sigma^2(1+\theta)^2,
$$
as
\begin{eqnarray*}
J &=& \sum_{j=-\infty}^{\infty}\gamma_j \\
&=& \gamma_0 + 2 \sum_{j=1}^{\infty}\gamma_j \\
&=& \gamma_0 + 2 \gamma_1 + 0\\
&=& \sigma^2(1 + \theta^2) + 2 \theta \sigma^2\\
&=& \sigma^2(1 + \theta^2 + 2\theta)\\
&=& \sigma^2(1 + \theta)^2
\end{eqnarray*}
We have $J=0$ for $\theta=-1$, so an $MA(1)$ with a unit root. This is a problem for example because the long-run variance is asymptotic variance of the sample mean,
$$
\sqrt{T}(\bar{Y}_T-\mu)\to_d N\Biggl(0,\sum_{j=-\infty}^{\infty}\gamma_j\Biggr),
$$
which is for instance used for standard errors - which should not be zero.
