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I have $\sum_{n\geq0}(2n)x^{3n} =0+2x^3+4x^6+6x^9+...$ , but i want to write this summation in terms of $x^n$ instead of $x^{3n}$ .How can i do it ?

I thought that if i can write $n/3$ in place of $n's$ , i can find somethings such that $$\sum_{n\geq0}(2n)x^{3n} =\sum_{n\geq0}(2n/3)x^{n}$$

However , in this case i obtain : $\sum_{n\geq0}(2n/3)x^{n}=0+(2/3)x+(4/3)x^2+2x^3 +...$ , but we should eliminate the terms $(2/3)x+(4/3)x^2$ etc ,because they are not the terms of $\sum_{n\geq0}(2n)x^{3n} =0+2x^3+4x^6+6x^9+...$

Can you help me ?

EDIT My question i was inspired from How to solve the $a_n$ in formal power series? , from @RobPratt's answer

2 Answers2

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Let $j = e^{\frac{2 i \pi}{3}}$ then : $$\forall n \in \mathbb{N}, 1 + j^n + j^{2 n} = \begin{cases} 3 & \text{if } 3 \mid n \\ 0 & \text{otherwise} \end{cases}$$ We deduce that : $$\sum_{n = 0}^{+\infty} 2 n x^{3 n} = \sum_{n = 0}^{+\infty} \dfrac{2}{9} n (1 + j^n + j^{2 n}) x^n$$

Essaidi
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  • very nice !!can you please explain your answer a little bit detailed , how reach the formula , why you used "j" etc –  Jun 12 '22 at 15:47
  • coz $j^3 = 1$ and $1 + j + j^2 = 0$. )if $n = 3 k$ then $1 + j^{3 k} + j^{6 k} = 1 + 1 + 1 = 3$. )if $n = 3 k + 1$ then $1 + j^{3 k + 1} + j^{6 k + 2} = 1 + j + j^2 = 0$. *) if $n = 3 k + 2$ then $1 + j^{3 k + 2} + j^{6 k + 4} = 1 + j^2 + j = 0$. – Essaidi Jun 12 '22 at 15:51
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Just write $\sum_{n\ge 1}c_nx^n$, where $c_n=0$ for $3\nmid n$ and $c_n=2n/3$ for $3\mid n$.

You cannot just write $n/3$ in place of $n$ because $n$ is also the summation index. So you need to introduce a function $c(n)$. It is given explicitly by $$ c(n)=\frac{2n}{9}(1+\omega^n+\omega^{2n}) $$ where $\omega$ is a primitive third root of unity.

Dietrich Burde
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