# Proving summation identities [duplicate]

How would one go about proving the following identities?

$$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i}{z_i-z_j} = \frac{n(n-1)}{2}$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^2}{z_i-z_j} = (n-1)\sum_{i=1}^n z_i$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^3}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^2+\sum_{i<j}^n z_i z_j$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^4}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^3+\sum_{i<j}^n z_i z_j(z_i+z_j)$$ $$\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^5}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^4+\sum_{i<j}^n z_i z_j(z_i^2+z_i z_j +z_j^2)$$

I see the obvious pattern here. The problem is that the algebra involving the summation is giving me some difficulty.

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## marked as duplicate by Peter Taylor, TMM, Davide Giraudo, Ayman Hourieh, JonathanJan 12 '14 at 17:35

What is $\large\left\{z_{i}\right\}$ ?. –  Felix Marin Jan 12 '14 at 2:33
Your identities are not correct. For $n=2$, your first identity becomes $\frac{z_1}{z_1-z_2}=1$. –  Phira Jan 12 '14 at 6:52
@Phira You're incorrect. For n=2, it would become $$\sum_{i=1}^2 \sum_{i\neq j}^2 \frac{z_i}{z_i-z_j} = \sum_{i=1}^2 \left(\frac{z_1}{z_1-z_2} + \frac{z_2}{z_2-z_1} \right)= \sum_{i=1}^2 1 = \frac{2(2-1)}{2} = 1$$ –  Millardo Peacecraft Jan 12 '14 at 7:03
@MillardoPeacecraft Dont you think that it is dishonest and pathetic to edit your question and then tell me that I was wrong to tell you that there was an error? –  Phira Jan 12 '14 at 8:26

$\sum_{i=1}^{n}\sum_{i \neq j}^n \frac{z_i}{z_i-z_j}$ $=\sum_{i=1}^{n}\sum_{i \neq j}^n \left(1+\frac{z_j}{z_i-z_j} \right)=n(n-1)-\sum_{i=1}^{n}\sum_{i \neq j}^n \frac{z_j}{z_i-z_j}.$
You can now use the same strategy to split $\frac{z_i^2}{z_i-z_j}=\frac{(z_i^2-z_j^2)+z_j^2}{z_i-z_j}=(z_i+z_j)+\frac{z_j^2}{z_i-z_j}$