If B is backshift operator, then how to calculate 1/(1 - B)? We have known that:
when |r|<1, (1−r)^-1 = 1 + r + r^2 + r^3 + r^4 + . . .
But for 1/(1-B), since the coefficient of B is 1, the power series 1 + B + B^2 + B^3 + . . . can not converge. In my opinion, we can not equal 1/(1-B) with that power series.
So how to calculate 1/(1-B)?
 A: I think that you are right. For instance, consider a simple AR(1) process:
\begin{align}
y_t&=\phi_1y_{t-1}+\epsilon_t \quad, \epsilon_t \sim WN(0,\sigma^2) \\
y_t-\phi_1y_{t-1}&=\epsilon_t \\
(1-\phi_1B)y_t&=\epsilon_t \\
\Phi(B)y_t&=\epsilon_t
\end{align}
Where $\Phi(B)=1-\phi_1B$ is the AR-polynomial. If $\vert \phi_1 \vert<1$, this polynomial is invertible, which means that
\begin{align}
\Phi(B)^{-1}=(1-\phi_1B)^{-1}=\frac{1}{1-\phi_1B}=1+\phi_1B+\phi_1B^2+\phi_1^3B^3+\dots
\end{align}
is finite. This is the reason why we can express a stationary AR(1) process as a MA($\infty$) process by simply inverting the AR-polynomial. I think that a simple random walk (the case $\phi_1=1$) corresponds to what you mean with $(1-B)^{-1}$. In this case it is not possible to invert the AR-polynomial, i.e. we cannot calculate $\Phi(B)^{-1}$. You can think of it as  $1+B+B^2+B^3+\dots$ is not finite.
Take a look at Hamilton's book "Time Series Analysis". The first three chapters deal with such type of questions.
