# 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??

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.
\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}
• Not stupid, you would have to try values for $n$. – Graham Hesketh Sep 25 '13 at 8:19