= Solution
The <unilateral shift operator> $R$ is not onto, since its outputs have first coordinate zero. The <left shift operator> $L$ is not injective, since $Le_1=0$. Both have <operator norm> one, and $LR=I$. Consequently
$$
L(R+A)=LR+LA=I+LA.
$$
If $\|A\|<1$, then $\|LA\|<1$, so the <Neumann series> makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would give
$$
L=(I+LA)(R+A)^{-1},
$$
a product of invertible operators, contradicting the nontrivial <kernel> of $L$. Thus
$$
\boxed{\|A\|<1\Longrightarrow R+A\text{ is not invertible}.}
$$
This is <norm stability of noninvertibility for the unilateral right shift>.
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