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Trying to prove that the formal power series with coefficients in a field $F$ is a commutative ring with identity. I have proved up to associativity of multiplication and am stuck on seeing how the indexes fall into place. The operation $\cdot$ is defined as

$$\sum_{n=0}^\infty a_nx^n \cdot \sum_{n=0}^\infty b_nx^n = \sum_{n=0}^\infty\bigg(\sum_{k=0}^n a_kb_{n-k}\bigg)x^n$$

I have the following

\begin{align*} \sum_{n=0}^\infty a_nx^n \cdot \bigg(\sum_{n=0}^\infty b_nx^n \cdot \sum_{n=0}^\infty c_nx^n\bigg)&=\sum_{n=0}^\infty a_nx^n \cdot \sum_{n=0}^\infty\bigg(\sum_{k=0}^n b_kc_{n-k}\bigg)x^n\\ &=\sum_{n=0}^\infty \bigg(\sum_{j=0}^na_j\bigg(\sum_{k=0}^{n-j}b_{k-j}c_{n-k-j}\bigg)\bigg)x^n\\ &=\sum_{n=0}^\infty \bigg(\sum_{j=0}^n\sum_{k=0}^{n-j}a_jb_{k-j}c_{n-k-j}\bigg)x^n\\ \end{align*}

I am not sure if the indexes here are correct and I am unsure how to proceed from here to prove associativity of multiplication.

Zev Chonoles
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kaiserphellos
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1 Answers1

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Almost right; it should be $$\begin{align*} \sum_{n=0}^\infty a_nx^n \cdot \bigg(\sum_{n=0}^\infty b_nx^n \cdot \sum_{n=0}^\infty c_nx^n\bigg)&=\sum_{n=0}^\infty a_nx^n \cdot \sum_{n=0}^\infty\bigg(\sum_{k=0}^n b_kc_{n-k}\bigg)x^n\\ &=\sum_{n=0}^\infty \bigg(\sum_{j=0}^na_j\bigg(\sum_{k=0}^{n-j}{\color{red}{b_{k}c_{(n-j)-k}}}\bigg)\bigg)x^n\\ &=\sum_{n=0}^\infty \bigg(\sum_{j=0}^n\sum_{k=0}^{n-j}a_j{\color{red}{b_{k}c_{n-j-k}}}\bigg)x^n \end{align*}$$ Now note that you are summing over all $a_jb_kc_m$ where $j+k+m=n$, and that this is symmetric in $j$, $k$, and $m$ (alternatively, do the same computation for the other order of association, and show you get the same such sum).

Zev Chonoles
  • 129,973