Double summation. I'm in the middle of an assignment, and I'm not looking for too much help, just more of a push in the right direction (as I haven't really encountered this in my mathematics courses before).
I'm looking at Weyl's dimension formula, and I have to prove a few things for gl(n)-modules. So, basically, we have to look at $\rho$ in the formula, with...
$$\rho = \frac{1}{2} (\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \varepsilon_{i} - \varepsilon_{j})$$
I understand that I'm trying to bring this to a singular sum, but I'm not quite sure where to even start. Do I need to manipulate the subscripts, particularly with $\varepsilon_{j}$?? Even in doing that, what does it do??
 A: Below shows how to manipulate the double sum into a single sum, apologies if it contains too much detail but it was difficult to explain without actually doing it.
To move from the first line to the second, in the left square bracket the inner sum is over $j$ but the index is $i$ consequently this just adds up the same term $n-i$ times. In the square bracket on the right is the non trivial part, to realise this relationship you can expand the terms by trying values for $n$ and then spot the pattern of how often each term appears.
To move from the second line to the third, arrange the left square bracket so that the summation limits match the right square bracket by pulling out unwanted terms, then combine the sums in the fourth line; you will then see that you can stick back in the loose term to form the fifth line.
$$\begin{aligned}  
\sum _{i=1}^{n-1} \left( \sum _{j=i+1}^{n}\epsilon_{{i}}-\epsilon_{{j}  } \right) &=\left[\sum_{i=1}^{n-1} \left( \sum _{j=i+1}^{n}\epsilon_{{i}} \right) \right]-\left[\sum_{i=1}^{n-1} \left( \sum_{j=i+1}^{n}\epsilon_{{j}}   \right)\right] \\
&=\left[\sum _{i=1}^{n-1} \left( n-i \right)\epsilon_{{i}}\right]-\left[\sum _{  i=2}^{n} \left( i-1 \right)\epsilon_{{i}}\right]\\
&=\left[\left( n-1 \right) \epsilon_{{1}}+\sum _{i=2}^{n} \left( n-i \right) \epsilon_{{i}}\right]-\left[\sum _{i=2}^{n} \left( i-1 \right) \epsilon_{{i}}\right]\\  &=\left( n-1 \right) \epsilon_{{1}}+\sum _{i=2}^{n} \left( n-2\,i+1 \right) \epsilon_{{i}}\\   
&=\sum _{i=1}^{n} \left( n-2\,i+1 \right) \epsilon_{{i}}\\     
\end{aligned}$$
