What is the correct notation of nested sigma (summation) signs? 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?
 A: 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 $$\sum_{(m_1,\dots,m_{n-1})\in S}m_{n-1}.$$
A: 
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: $(0,1,2,\ldots,n-2,n-1)$.
