Changing the order of summation in a triple sum I am trying to simplify an expression that came up when I was trying to calculate the energy of a multiple pendulum system.
The expression I have is
$$\sum_{k=1}^N \sum_{i=1}^k \sum_{j=1}^k m_k v_i v_j \cos(\theta_i-\theta_j)$$
but I want to change the order of summation so that the sum over $k$ is now on the inside. After a lot of trial and error, I seem to have found that
$$\sum_{k=1}^N \sum_{i=1}^k \sum_{j=1}^k m_k v_i v_j \cos(\theta_i-\theta_j)=\sum_{i=1}^N \sum_{j=1}^N \sum_{k=\max(i,j)}^N m_k v_i v_j \cos(\theta_i-\theta_j)$$
works, but I have no idea where this result comes from or how one would prove it. Any help would be appreciated, thanks!
 A: Let us examine the set over which the summation is taken. The values of $i$ and $j$ are independent of each other, but both depends on $k$. 
\begin{align}
1\leq& k\leq N\\
1\leq &i\leq k\\
1\leq &j\leq k
\end{align}
Next, we change the bounds on $k,i,j$ such that $k$ is bounded in terms of $i$ and $j$, and bounds on $i$ and $j$ are independent of $k$. It is easy to see that $1\leq i,j\leq N$. Further, we have
\begin{align}
1\leq &i\leq k \text{ and } 1\leq k\leq N \implies i\leq k\leq N\\
1\leq &j\leq k \text{ and } 1\leq k\leq N \implies j\leq k\leq N.
\end{align}
Thus, we get
\begin{equation}
\max\{i,j\}\leq k\leq N.
\end{equation}
Therefore, the result. If $i$ and $j$ were dependent, we have to do this cyclically, i.e., first find bound on $i$ which is independent of $j$ and $k$. Then, bound $j$ interms of $i$, and finally, bound $k$ using $i$ and $j$.
A: We  can  write     the   triple   sum as
\begin{align*}
\sum_{k=1}^N \sum_{i=1}^k \sum_{j=1}^k m_k v_i v_j \cos(\theta_i-\theta_j)
&=\sum_{\color{blue}{1\leq i,j\leq k\leq N} }m_k v_i v_j \cos(\theta_i-\theta_j)\\
&=\sum_{i=1}^N \sum_{j=1}^N\sum_{k=\max\{i,j\}}^N  m_k v_i v_j \cos(\theta_i-\theta_j)
\end{align*}
The blue marked index range in the middle sum clearly indicates that the left and right representations are identical. There is nothing left to prove.
