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$\{ x \in \mathbb{C}$ | $x^{2n+1} = 1$ , $x \neq 1\}$

Compute $\displaystyle{\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}}$

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1  
What if 2n+1 has a divisor m such that $x^m = 1$?Then you get a zero divisor in the sum. – marty cohen Mar 26 '12 at 2:18
    
You call this precalculus? – Robert Israel Mar 26 '12 at 2:45
up vote 3 down vote accepted

Since

$$(x^{\small{2n+1}}-1) = (x-1)(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0$$

Therefore $$(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0 \hspace{4pt} ({\text{since}} \hspace{4pt} x \neq 1) \tag{1}$$

$$ \begin{align*} S = \sum_{i=1}^{\small{2n}} \frac{x^{\small{2i}}}{x^i-1} &= \sum_{i=1}^{\small{2n}} x^i + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1}\\ &= -1 + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \hspace{12pt} {\text{from}} \hspace{4pt} (1) \\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n+1}}}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \tag{2} \end{align*} $$

Since $x^{\small{2n+1}} = 1$

$$ \frac{x^{\small{n+i}}}{x^{\small{n+i}}-1} = \frac{-1}{x^{\small{n+1-i}}-1} $$

Therefore $(2)$ can be simplified to

$$ \begin{align*} & -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n}}}^{\small{1}} \frac{-1}{x^{\small{n+1-i}}-1}\\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{1}}}^{\small{n}} \frac{-1}{x^{\small{i}}-1}\\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}-1}{x^i-1} \\ &= n-1 \end{align*} $$

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