TITLE: Nilpotent elements in power series over a Noetherian ring
QUESTION [8 upvotes]: Suppose $A$ is a Noetherian ring. Let $$f(x)=\sum_{i=0}^{\infty} a_i x^i\in A[[x]].$$ If $a_i$ are all nilpotent, then $f$ is nilpotent. How to prove it?

REPLY [8 votes]: Hint: the sequence of ideals $(a_0), (a_0) + (a_1), (a_0) + (a_1) + (a_2), \ldots$ stabilizes after finitely many steps. 
Edit (years later): It was asked in a comment how this implies that $f$ is nilpotent. 
Let $I = (a_0) + \ldots + (a_j)$ be the ideal to which the sequence stabilizes. Thus each $a_n$ can be written in the form $\sum_{i=0}^j a_i r_{i, n}$, so that 
$$f(x) = \sum_{n \geq 0} a_n x^n = \sum_{n \geq 0} \left(\sum_{i = 0}^j a_i r_{i, n}\right) x^n = \sum_{i=0}^j a_i f_i(x)$$ 
where $f_i(x) = \sum_{n \geq 0} r_{i, n} x^n$. Each $a_i f_i(x)$ is nilpotent since $a_i$ is, and a sum of finitely many nilpotent elements is nilpotent by a simple inductive argument based on the binomial expansion $(a + b)^n = \sum_{j=0}^n \binom{n}{j} a^j b^{n-j}$. This completes the proof.