$$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n} 1/2 = \sum_{i=1}^{n-1} (n-1) (1/2) $$ We remove the first summation because it just sums all $1/2$ in $n-1$ times. So we get $(n-1)(1/2)$. continue: $$ \sum_{i=1}^{n-1} (n-1) (1/2) = (n-1)(n-1)(1/2)$$ Is this correct? Because the answer should be something like $n(n-1)/4$.
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You cannot remove the outer summation, since the inner one depends on it. What you can do is simplify the inner summation first, and then the outer summation. The inner summation is summing only $\frac 12$, but the number of halves depends on $i$, so you have to be careful in evaluation. – Sarvesh Ravichandran Iyer Mar 06 '17 at 12:50
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It's not correct. The inner summation sums $1/2$ not $n-1$ times, but $n-i$ times. Therefore you should write $\sum_{j=i+1}^n 1/2 = (n-i)/2$. Doing this, we get $$\sum_{i=1}^{n-1} \sum_{j=i+1}^n 1/2 = \sum_{i=1}^{n-1} (n-i)1/2 = 1/2 \cdot \sum_{i=1}^{n-1} (n-i) = 1/2 \cdot n(n-1)/2 = n(n-1)/4.$$
Integral
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