Lemma 1   Let $A \subset B$ be ideals of a ring $R$ If $A$ is nil and $B/A$ is nil, then $B$ is nil. If $A$ is nilpotent and $B/A$ is nilpotent, then $B$ is nilpotent.

Proof. Suppose that $A$ and $B/A$ are nil. Let $x \in B$ Then $x^n \in A$ for some $n$ since $B/A$ is nil. But $A$ is nil, so there is an $m$ such that $x^{nm} = (x^n)^m = 0$ Thus $B$ is nil.

Suppose that $A$ and $B/A$ are nilpotent. Then there are natural numbers $n$ and $m$ such that $A^m = 0$ and $B^n \subseteq A$ Therefore, $B^{nm} = 0$

Lemma 2   The sum of an arbitrary family of nil ideals is nil.

Proof. Let $R$ be a ring, and let $\famf$ be a family of nil ideals of $R$ Let $S = \sum_{I \in \famf} I$ We must show that there is an $n$ with $x^n = 0$ for every $x \in S$ Now, any such $x$ is actually in a sum of only finitely many of the ideals in $\famf$ So it suffices to prove the lemma in the case that $\famf$ is finite. By induction, it is enough to show that the sum of two nil ideals is nil.

Let $A$ and $B$ be nil ideals of a ring $R$ Then $A \subset A + B$ and $A+B/A \cong B/(A \cap B)$ which is nil. So by the first lemma, $A + B$ is nil.

Lemma 3   The sum of a finite family of nilpotent left or right ideals is nilpotent.

Proof. We prove this for right ideals. Again, by induction, it suffices to prove it for the case of two right ideals.

Let $A$ and $B$ be nilpotent right ideals of a ring $R$ Then there are natural numbers $n$ and $m$ such that $A^n = 0$ and $b^m = 0$

Let $k = n + m -1$ Let $z_1, z_2, \dots, z_k$ be elements of $A + B$ We may write $z_i = a_i + b_i$ for each $i$ with $a_i \in A$ and $b_i \in B$ If we expand the product $z_1z_2 \dotsm z_k$ we get a sum of terms of the form $x_1x_2 \dots x_k$ where each $x_i \in \{ a_i, b_i \}$

Consider one of these terms $x_1x_2 \dotsm x_k$ Then by our choice of $k$ it must contain at least $n$ of the $a_i$ s or at least $m$ of the $b_i$ s. Without loss of generality, assume the former. So there are indices $i_1 < i_2 < \dots < i_n$ with $x_{i_j} \in A$ for each $j$ For $1 \le j \le n-1$ define $y_j = x_{i_j} x_{i_j+1} \dotsm x_{i_{j+1}-1}$ and define $y_n = x_{i_n}x_{i_n+1} \dotsm x_k$ Since $A$ is a right ideal, $y_j \in A$

Then $x_1x_2 \dotsm x_k = x_1 x_2 \dotsm x_{i_1-1} y_1 y_2 \dotsm y_n \in x_1 x_2 \dotsm x_{i_1-1} A^n = 0$

This is true for all choices of the $x_i$ and so $z_1 z_2 \dotsm z_k = 0$ But this says that $(A+B)^k = 0$