= Solution
{c}
\b[Yes: $R[[x]]$ is Noetherian.] We prove <Noetherianity of a formal power series ring> directly, using only the equivalence in part (a). Let $A=R[[x]]$, a <formal power series ring>, and fix an <ideal> $I\subseteq A$. For $n\geq0$, let $C_n$ be the set of coefficients of $x^n$ in elements of $I\cap x^nA$. It is an <ideal> of $R$, because coefficient extraction is additive and respects multiplication by constants. Multiplication by $x$ gives
$$
C_0\subseteq C_1\subseteq C_2\subseteq\cdots.
$$
Since $R$ is <Noetherian>, there is an $N$ with $C_n=C_N$ for all $n\geq N$. Each $C_n$, for $0\leq n\leq N$, has a finite <generating set of an ideal> $a_{n1},\ldots,a_{nr_n}$. Choose $f_{nj}\in I\cap x^nA$ whose coefficient of $x^n$ is $a_{nj}$. If $C_n=0$, use no generators at that index.
We claim that these finitely many $f_{nj}$ generate $I$ over $A$. Given $f\in I$, start with $g_0=f$ and recursively construct $g_n\in I\cap x^nA$. For $n<N$, express the coefficient of $x^n$ in $g_n$ as $\sum_jc_{nj}a_{nj}$, with $c_{nj}\in R$, and subtract $\sum_jc_{nj}f_{nj}$ to obtain $g_{n+1}\in I\cap x^{n+1}A$. For $n\geq N$, use $C_n=C_N$ and subtract
$$
\sum_jc_{nj}x^{n-N}f_{Nj}
$$
to cancel that coefficient. Every step involves a finite sum and leaves a residual still in $I$.
For each $j$, define the <formal power series>
$$
h_j(x)=\sum_{n=N}^{\infty}c_{nj}x^{n-N}\in A.
$$
The residual after the $n$th cancellation lies in $x^{n+1}A$, so comparison of each coefficient gives the exact identity
$$
f=\sum_{n=0}^{N-1}\sum_jc_{nj}f_{nj}+\sum_jh_j(x)f_{Nj}.
$$
The right side is a finite $A$-linear combination of the chosen generators. No assertion that arbitrary <ideals> are closed under limits is being used: the equality is verified coefficient by coefficient, and the multipliers $h_j$ belong to $A$. Thus every <ideal> of $A$ is finitely generated, proving
$$
\boxed{R\text{ Noetherian}\ \Longrightarrow\ R[[x]]\text{ Noetherian}.}
$$
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