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If $a_{0},a_{1},a_{2},a_{3}....a_{n-1}$ are the nth roots of unity, then the value of $$\sum_{i=0}^{n-1}{\left(\frac{a_{i}}{3-a_{i}}\right)}.$$

I tried expressing $a_{i}$ in euler form as $e^{\frac{2k\pi i}{n}}$ but I am stuck after it.

Edit: It is a question from JEE examination.

Ѕᴀᴀᴅ
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Equation_Charmer
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    This problem looks quite engaging but I can't read your mathematical expressions. Try $LaTeX$ ! – Robert Lewis Mar 07 '20 at 02:07
  • Can you use complex integration and residues? – robjohn Mar 07 '20 at 04:22
  • I am just a class 11 student in India. I dont know all these – Equation_Charmer Mar 07 '20 at 04:25
  • Try using the Taylor series $\frac{x}{1-x}=\sum\limits_{m=1}^\infty x^m$ and then $\sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=n,[n\mid m]$ (Iverson brackets). Add your attempt to your question (or write an answer). – robjohn Mar 07 '20 at 06:09
  • I even dont know these. I have already added my attempt in the question. I could proceed only to one step. – Equation_Charmer Mar 07 '20 at 16:58
  • If you haven't covered these in class, then I am surprised that this question was given. Perhaps this wasn't given in class; in any case, please add some context, then we would know what you can use and what you can't. I have written both a complex integration and a series based answer using my hint above, only to be told you can't use those methods. – robjohn Mar 09 '20 at 21:01
  • Are you familiar with the sums of geometric series? Then you should know that $\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k$. If you know that $\frac{1-x^n}{1-x}=\sum\limits_{k=0}^{n-1}x^k$, from which you can establish $\sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=n,[n\mid m]$. – robjohn Mar 10 '20 at 03:54
  • I accepted it by mistake. And do post the answer with school mathematic concepts – Equation_Charmer Mar 11 '20 at 11:27
  • See https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity/1909366#1909366 and https://math.stackexchange.com/questions/2258846/calculate-product-of-n-th-roots-of-unity – lab bhattacharjee Apr 17 '20 at 05:21

2 Answers2

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$$ \begin{align} \sum_{k=0}^{n-1}\frac{e^{\frac{2\pi ik}{n}}}{3-e^{\frac{2\pi ik}{n}}} &=\sum_{k=0}^{n-1}\frac{\frac13e^{\frac{2\pi ik}{n}}}{1-\frac13e^{\frac{2\pi ik}{n}}}\tag1\\ &=\sum_{m=1}^\infty\sum_{k=0}^{n-1}\frac1{3^m}e^{\frac{2\pi ikm}{n}}\tag2\\ &=\sum_{m=1}^\infty\frac1{3^m}n\,[n\mid m]\tag3\\ &=\sum_{j=1}^\infty\frac1{3^{jn}}n\tag4\\[6pt] &=\frac{n}{3^n-1}\tag5 \end{align} $$ Explanation:
$(1)$: cancel $3$ in numerator and denominator
$(2)$: use the Taylor series $\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k$
$(3)$: $\sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=n\,[n\mid m]$ (Iverson brackets)
$(4)$: select the terms where $m=jn$
$(5)$: sum the geometric series

Explanation of step $\boldsymbol{(3)}$

Note that the sum of the following finite geometric series $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=\frac{e^{\frac{2\pi inm}{n}}-1}{e^{\frac{2\pi im}{n}}-1}=\frac0{e^{\frac{2\pi im}{n}}-1}\tag6 $$ shows that if $n\nmid m$, $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=0\tag7 $$ It is pretty simple to see that if $n\mid m$, $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=\sum\limits_{k=0}^{n-1}1=n\tag8 $$

robjohn
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This is only a hint,

$$ \begin{aligned} \sum_{i=0}^{n-1}{\left(\frac{a_{i}}{3-a_{i}}\right)}+n&=\sum_{i=0}^{n-1}{\left(\frac{1}{1-\frac{a_{i}}{3}}\right)}\\ &=\frac{\sum_{i=0}^{n-1}{\left(\prod_{j\neq i}{\left(1-\frac{a_{j}}{3}\right)}\right)}}{\prod_{i=0}^{n-1}{\left(1-\frac{a_{i}}{3}\right)}} \end{aligned} $$

Then You can use Vieta's formula.

acat3
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