T-nilpotency and quasinilpotency of ideals Let $R$ be a commutative ring and let $\mathfrak{a}\subseteq R$ be an ideal. The ideal $\mathfrak{a}$ is called T-nilpotent if for every sequence $(r_i)_{i\in\mathbb{N}}$ in $\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $\prod_{i=0}^nr_i=0$, and quasinilpotent if there exists $n\in\mathbb{N}$ such that for every $r\in\mathfrak{a}$ we have $r^n=0$.
If $\mathfrak{a}$ is nilpotent (i.e., there exists $n\in\mathbb{N}$ with $\mathfrak{a}^n=0$), then it is T-nilpotent and quasinilpotent. If $\mathfrak{a}$ is T-nilpotent or quasinilpotent, then it is nil (i.e., for every $r\in\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $r^n=0$). 
Conversely, a nil ideal need not be T-nilpotent or quasinilpotent, a T-nilpotent ideal need not be nilpotent, and a quasinilpotent ideal need not be nilpotent.
I guess that a T-nilpotent and quasinilpotent ideal need not be nilpotent either. However, I was not able to come up with an example of such an ideal. Thus:

What is an example of an ideal in a commutative ring that is T-nilpotent, quasinilpotent, but not nilpotent?

 A: No need to specify the ambient ring since every commutative associative ring $I$ (possibly without unit) is an ideal in a ring (namely $I\oplus\mathbf{Z}$ with multiplicative law $(a_1,n_1)(a_2,n_2)=(a_1a_2+n_1a_2+n_2a_1,n_1n_2)$). 
Now call a X-ring a ring (possibly without unit) satisfying the axioms: associative, commutative, $x+x=x^2=0$ for all $x$.
So the free X-ring $R_I$ on the generators $(x_i)_{i\in I}$ has the basis over the field on 2 elements: $(x_J)_{J\in 2^I_0}$, where $2^I_0$ denotes the set of nonempty finite subsets of $I$, and $x_J=\prod_{j\in J}x_j$.
If $I$ is finite, then $R_I$ is finite and nilpotent (of length $\le 2^{|I|}$ but I'm lazy to get the best bound).
Then $R_I$ is not T-nilpotent when $I$ is infinite. But now let $I$ denote the positive integers, and define $S$ as the quotient modding out by $(x_ix_j)$ whenever $2i<j$. It has the basis $(x_J)$ where $J$ ranges over nonzero finite subset not containing any pair of the form $\{i,j\}$ with $2i<j$.
(a) $S$ is quasinilpotent (with $n=2$ as any X-ring) 
(b) $S$ is not nilpotent (because $x_n\dots x_{2n}\neq 0$ for every $n$)
(c) $S$ is T-nilpotent: indeed, consider a sequence $(y_n)$. Write $y_1=\sum_{k=1}^mx_{J_k}$, with $J_k$ nonempty, pairwise distinct and not containing any pair $\{i,j\}$ with $2i<j$; let us show $\prod_{i=1}^p y_k=0$ for large $p$. We can assume $m>0$ since otherwise $y_1=0$. Define $n=\max\bigcup J_k$. Let $\pi$ map $x_J$ to itself if $J\subset\{1,\dots,2n\}$ and map $x_J$ to 0 otherwise, and extend it by linearity; this is a projection, and $y_1y=y_1\pi(y)$ for all $y\in S$. Hence $y_1\dots y_p=y_1\pi(y_2)\dots \pi(y_p)$. Since this product lies in the (finite) subring generated by $x_1,\dots,x_{2n}$, which is nilpotent, it is 0 for large $p$ (say $p\ge 4^n$).
A: Based on Yves answer above, let me give another example (or maybe another description of his example - I am not sure about this).
Let $K$ be a field of characteristic $2$, let $$R=K[(X_i)_{i\in\Bbb N}]/\langle\{X_i^2\mid i\in\Bbb N\}\cup\{X_iX_j\mid i,j\in\Bbb N,2i<j\}\rangle,$$ denote for $i\in\Bbb N$ by $Y_i$ the canonical image of $X_i$ in $R$, and let $\mathfrak a=\langle Y_i\mid i\in\Bbb N\rangle$.
$\mathfrak a$ is not nilpotent: If $\mathfrak a^n=0$, then $Y_n\cdots Y_{2n}=0$, hence there are $i,j\in[n,2n]$ with $2i<j$ - contradiction.
$\mathfrak a$ is quasinilpotent: An element of $\mathfrak a$ has the form $\sum_{i=0}^ra_iY_i$ with $a_i\in R$, and $(\sum_{i=0}^ra_iY_i)^2=\sum_{i=0}^ra_i^2Y_i^2=0$.
$\mathfrak a$ is T-nilpotent: Assume there is a sequence $(r_i)_{i\in\Bbb N}$ in $\mathfrak a$ such that for every $n\in\Bbb N$ we have $\prod_{i=0}^nr_i\neq 0$. Then, there exists such a sequence in which every $r_i$ is a monomial in $(Y_i)_{i\in\Bbb N}$. Without loss of generality we can thus suppose $r_i=Y_{j_i}$ for every $i\in\Bbb N$, where $(j_i)_{i\in\Bbb N}$ is an injective family in $\Bbb N$. Injectivity of this family implies that there exists $k\in\Bbb N$ with $j_k>2j_0$, and thus we get the contradiction $\prod_{i=0}^kr_i=0$. 
