# Calculate a sum involving nth root of unity

Calculate $$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}$$ Where $\epsilon$ is nth root of unity. There is a hint that says: multiply by $(1-\epsilon)$

Doing this multiplication I get: $$1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}-n$$ The answer is: $-\frac{n}{1-\epsilon}$if $\epsilon\neq1$ and $\frac{n(n+1)}{2}$ if $\epsilon=1$. I can't see how to get this result...

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Is $\epsilon$ a complex number? If not you should change your tag. –  Mercy Oct 3 '12 at 11:36
@Mercy , it says in the question: "where $\epsilon$ is nth root of unity" –  Mykolas Oct 3 '12 at 11:38
sorry, I didn't pay attention :D –  Mercy Oct 3 '12 at 12:44
Hint: multiply by $(1-\epsilon)$ again... –  Erick Wong Oct 3 '12 at 13:31
A formula for $\sum\limits_{r = 1}^{d} r x^{r}$ si given in this answer. (It is more complicated, since it is a formula for general $x$; without using $x^d=1$). –  Martin Sleziak Oct 3 '12 at 13:48

If $\epsilon=1$, $$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}=\sum_{k=1}^nk=\frac{n(n+1)}2$$ by the familiar formula for the sum of the first $n$ positive integers (or more generally by the formula for the sum of a finite arithmetic series).

If $\epsilon\ne 1$, you’ve shown that

$$(1-\epsilon)(1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1})=\sum_{k=0}^{n-1}\epsilon^k-n=\frac{1-\epsilon^n}{1-\epsilon}-n=-n\;,$$

where I’ve used the formula for the sum of a finite geometric series and the fact that $\epsilon^n=1$. It follows immediately that

$$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}=-\frac{n}{1-\epsilon}$$ in this case.

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Thankyou @Brian M. Scott –  Mykolas Oct 3 '12 at 11:41
@Mykolas: You’re welcome. –  Brian M. Scott Oct 3 '12 at 11:42

$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1} = f'(\epsilon)$, where $f(x)=1+x+x^2+\cdots+x^n= \dfrac{x^{n+1}-1}{x-1}$.

Now $f'(x)=\dfrac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2}$ and so $f'(\epsilon)=\dfrac{(n+1)(\epsilon-1)-(\epsilon-1)}{(\epsilon-1)^2}=\dfrac{n}{\epsilon-1}$, if $\epsilon\neq 1$.

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Of course, when $\epsilon=1$, \begin{align} &1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1}\\ &=1+2+2+4+\dots+n\\ &=\frac{n(n+1)}{2} \end{align} When $\epsilon\ne1$ \left. \begin{align} 1+\epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{1-\epsilon^n}{1-\epsilon}\\ \epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon-\epsilon^n}{1-\epsilon}\\ \epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^2-\epsilon^n}{1-\epsilon}\\ \epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^3-\epsilon^n}{1-\epsilon}\\ &\vdots\\ \epsilon^{n-1}&=\frac{\epsilon^{n-1}-\epsilon^n}{1-\epsilon}\\ \end{align} \right\}\mbox{ n equations} Summing these $n$ equations yields \begin{align} 1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1} &=\frac{\frac{1-\epsilon^n}{1-\epsilon}-n\epsilon^n}{1-\epsilon}\\ &=\frac{n}{\epsilon-1} \end{align}

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$1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}$ is a geometric series and its summation can be computed as $$\frac{1-\epsilon^n}{1-\epsilon}$$ for $\epsilon\neq 1$.

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