Sum of two right T-nilpotent right ideals 
Is it true that the sum of two left $T$-nilpotent left ideals is again left $T$-nilpotent? 

If $A$ and $B$ are two left $T$-nilpotent left ideals and $x_1,x_2,\dots$ belong to $A+B$, then each $x_j$ equals $a_{j}+b_{j}$ for some $a_j\in A,b_j\in B$. Now, we are looking for some integer $n$ with $x_1x_2\cdots x_n=0$ knowing that there exist integers $s$ and $t$ with $a_1a_2\cdots a_s=0$ and $b_1b_2\cdots b_t=0$.
If the answer is in the affirmative, one could deduce that any arbitrary sum of left $T$-nilpotent left ideals is again a left $T$-nilpotent left ideal.
 A: The usual definition of left $T$-nilpotence for a 1 or 2-sided
ideal $I$ of $R$ is that in
any sequence $(a_1, a_2, \ldots)$, with $a_i\in I$, there is an $n$ such that
$a_1a_2\cdots a_n = 0$.
The usual definition of right $T$-nilpotence for a 1 or 2-sided 
ideal $I$ of $R$ is that in
any sequence $(a_1, a_2, \ldots)$, with $a_i\in I$, there is an $n$ such that
$a_n\cdots a_1 = 0$.
I argue that the sum of two left $T$-nilpotent left ideals is left $T$-nilpotent.
I will argue the contrapositive, so assume that 
$A, B$ are left ideals of $R$ such that $A+B$
is not left $T$-nilpotent. Then there exists an infinite sequence 
$(a_1+b_1, a_2+b_2, \ldots)$
with $a_i\in A, b_i\in B$ such that
$$
(a_1+b_1)(a_2+b_2)\cdots (a_n+b_n)\neq 0
$$
for every $n$. Multiplying this out, it follows that, for every $n$,
there is a nonzero product of the form $c_1c_2\cdots c_n$
where the symbol $c$ is either $a$ or $b$.
Record this fact by collecting all such sequences
$(c_1,\ldots,c_n)$ into a set $G$, the "good sequences".
(E.g., $(c_1,c_2,c_3)$ could be $(a_1,a_2,a_3)$ or 
$(a_1,a_2,b_3)$ or $(a_1,b_2,a_3)$ or ETC, and it is "good" if the
product $c_1c_2c_3$ is not zero.)
Order the set of good sequences by saying that $\sigma\leq \tau$
for $\sigma,\tau\in G$ if $\sigma$ is an initial segment of $\tau$.
Since there are arbitrarily long good sequences, the set $G$
is an infinite tree under this order. It also finitely branching,
which means that if $\sigma\in G$ is
an $n$-term good sequence, then there are finitely many
ways to extend $\sigma$ to an $(n+1)$-term good sequence.
(In fact, there are at most 2 ways to extend $(c_1,\ldots,c_n)$, namely $(c_1,\ldots,c_n,a_{n+1})$ or $(c_1,\ldots,c_n,b_{n+1})$.)
Konig's Tree Lemma proves that there is an infinite branch
in the tree of good sequences. This means that there is a sequence
$\sigma_1, \sigma_2, \ldots$ such that $\sigma_{i+1}$
extends $\sigma_i$ by one term for each $i$. These
$\sigma_i$'s are partial approximations to an infinite
sequence $(c_1,c_2,c_3,\ldots)$ of elements in $A\cup B$
such that $c_1c_2\cdots c_n \neq 0$ for all $n$.
Now either infinitely many of the $c$'s lie in $A$ or
infinitely many lie in $B$ (or both). Let's say
that there exists an infinite sequence  $i_1 < i_2 < \cdots$
of subscripts such that $c_{i_j}\in A$ for all $j$. Then
by grouping terms and multiplying we get a sequence
$$
(d_1,d_2,d_3.\ldots) =
(\underline{c_1c_2\cdots c_{i_1}}, 
\underline{c_{i_{1}+1}\cdots c_{i_2}}, 
\underline{c_{i_{2}+1}\cdots c_{i_3}},
\ldots ),
$$
which is an infinite sequence of elements of $A$ where the product
of any initial segment is not zero.
This contradicts the left $T$-nilpotence of $A$.
If we were dealing with right ideals instead of left ideals, we would argue in exactly the same way, but group them differently at the end, namely as
$$
(\underline{c_{i_1}\cdots c_{i_2-1}}, 
\underline{c_{i_{2}}\cdots c_{i_3-1}},
\underline{c_{i_{3}}\cdots c_{i_4-1}},
\ldots).
$$
Each underlined term belongs to $A$, and if we multiply the elements in any initial segment we must get a nonzero value, since premultiplying the product with 
$c_1\cdots c_{{i_1}-1}$ yields a nonzero value. Thus the sum of any two left ideals or any two right ideals which are both left $T$-nilpotent is again left $T$-nilpotent.
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