Let $A$ be a commutative ring with $1$ and consider the ring of formal power series $A[[X]]$. If $I \subseteq A$ is an ideal, let $I[[X]]$ denote the set of power series with coefficients in $I$. This is an ideal; it is the kernel of the reduction homomorphism $A[[X]] \to (A/I)[[X]]$. Let $IA[[X]]$ denote the extended ideal.
Matsumura writes the following in his Commutative Ring Theory (p. 5):
...if $I$ is not finitely generated then $I[[X]]$ is bigger than $IA[[X]]$.
I believe if $I$ has a countably infinite set of generators, say $I=(a_0,a_1,\ldots)$, then the following will suffice as a proof: If $f=a_0 + a_1 X + a_2 X^2 + \cdots$, then $f$ is not in the exteneded ideal. Indeed, if it were, then $f = \sum_{i=1}^m b_i g_i$ for some $b_i \in I$ and $g_i \in A[[X]]$. But then $(a_0,a_1,\ldots) \subseteq (b_1,b_2,\ldots,b_m)$, so $I$ is finitely generated.
What is the proof for a general non-countably-generated ideal? My ideas aren't leading anywhere.
$I[[X]]=I A[[X]]$ holds if and only if for every at most countable subset $S \subseteq I$ there is some finite subset $B \subseteq I$ such that $(S) \subseteq (B)$. Clearly this holds when $I$ is finitely generated. But the converse is not true. There are examples with $|I|=\aleph_1$, the smallest uncountable cardinality. I think $A=k[t,\{X_{\alpha}\}_{\alpha<\aleph_1}]/(X_\alpha=t X_{\beta})_{\alpha<\beta}$ and $I=(X_{\alpha})_{\alpha<\aleph_1}$ should work.