How would an expression that looks like this be written?
$$\underbrace{\sum_{m_1=0}^{n-1}\sum_{m_2=0}^{m_1-1}\dots\sum_{m_{n-2}=0}^{m_{n-3}-1}\sum_{m_{n-1}=0}^{m_{n-2}-1}m_{n-1}}_{n-1\sum-symbols}$$
What is the correct notation for nested sigma (summation) symbols of this nature? Is this the most efficiently this equation can be written out?
On
We can considerably simplify this multiple sum since we obtain for $n\geq 1$:
\begin{align*} \color{blue}{\sum_{m_1=0}^{n-1}}&\color{blue}{\sum_{m_2=0}^{m_1-1}\dots\sum_{m_{n-2}=0}^{m_{n-3}-1}\sum_{m_{n-1}=0}^{m_{n-2}-1}m_{n-1}}\\ &=\sum_{m_1=0}^{n-1}\sum_{m_2=0}^{m_1-1}\dots \sum_{m_{n-2}=0}^{m_{n-3}-1}\sum_{m_{n-1}=0}^{m_{n-2}-1}\sum_{m_{n}=0}^{m_{n-1}-1}1\tag{1}\\ &=\sum_{0\leq m_{n}<m_{n-1}<\cdots<m_2<m_1\leq n-1}1\tag{2}\\ &\,\,\color{blue}{=1}\tag{3} \end{align*}
Comment:
In (1) we write $m_{n-1}$ as sum: $m_{n-1} =\sum_{m_n=0}^{m_{n-1}-1}1$.
In (2) we use another typical notation and write the index range as inequality chain.
In (3) we observe the index range contains exactly one $n$-tuple: $(0,1,2,\ldots,n-2,n-1)$.
One can write more compactly $$\sum_{\substack{0\le m_1<n\\0\le m_2<m_1\\\vdots\\0\le m_{n-2}<m_{n-3}\\0\le m_{n-1}<m_{n-2}}}m_{n-1}.$$ Personally, I would first define a set $$S=\{(m_1,\dots,m_{n-1})\in\mathbb N^{n-1}:n>m_1>m_2>\dots>m_{n-2}>m_{n-1}\ge0\}$$ and then write $$\sum_{(m_1,\dots,m_{n-1})\in S}m_{n-1}.$$