Multiplying power formal series when reindexing is required? Normally, when I multiply the two following series:
$$\sum_{n\ge0}a_nx^n$$
$$\sum_{n\ge0}b_nx^n$$
I have:
$$\sum_{n\ge0}(\sum_{i=0}^na_ib_{n-i})x^n$$
Which is nice and intuitive. However, what if the exponents of x, or the indices don't match up? For example, what if I wanted to multiply these two:
$$\sum_{n\ge0}a_nx^{100n}$$
$$\sum_{n\ge0}b_nx^n$$
If I were to try to do this I'd reindex the second series to:
$$\sum_{\frac{n}{100}\ge0}b_nx^{100n}$$
But then the indices wouldn't line up. Am I missing something?
 A: We have
\begin{eqnarray*}
\left( \sum_{n=1}^{ \infty} a_n x^{100n} \right)  \left(\sum_{  m=1}^{\infty }b_m x^m\right) = \sum \sum a_n b_m x^{ \color{red}{100n+m}}.
\end{eqnarray*}
So to collect "like" $x$ terms, we need to list solutions of $100n+m=N$ ... this will give
\begin{eqnarray*}
 \sum_{N=1}^{\infty}  \left(\sum_{m \mid m+100n=N} a_n b_m \right) x^{ \color{red}{N}}.
\end{eqnarray*}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\left.\begin{array}{r}
\ds{\sum_{n = 0}^{\infty}a_{n}x^{\mu n}}
\\[2mm]
\ds{\sum_{n = 0}^{\infty}b_{n}x^{\nu n}} 
\end{array}\right\}
\implies
\left\{\begin{array}{rcl}
\ds{\sum_{j = 0}^{\infty}a_{j}\, x^{\mu j}\sum_{k = 0}^{\infty}b_{k}\,x^{\nu k}} & \ds{=} &
\ds{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\, b_{k}\sum_{n = 0}^{\infty}
\delta_{n,\mu j + \nu k}\,\,\,x^{n}}
\\[2mm] & \ds{=} &
\ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\,b_{k}\,
\delta_{n,\mu j + \nu k}}}x^{n}
\\[2mm] & \ds{=} &
\bbx{\ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}a_{j}
\,b_{\pars{n - \mu j}/\nu}\,\,\,
\bracks{{n - \mu j \over \nu} \in \mathbb{N}_{\geq 0}}}}x^{n}} \\ &&
\end{array}\right.
\end{align}
A: We can nicely see the similarities when adding an intermediate step.

*

*On the one hand we have
\begin{align*}
\sum_{k=0}^\infty a_kx^k\sum_{l=0}^\infty b_lx^l
&=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n
=\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{n}} a_kb_{n-k} x^n\\
\end{align*}


*On the other hand we have
\begin{align*}
\sum_{k=0}^\infty a_kx^{100k}\sum_{l=0}^\infty b_lx^l
&=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{100k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n
=\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{\left\lfloor\frac{n}{100}\right\rfloor}}a_kb_{n-100k} x^n\\
\end{align*}
Comment:

*

*In the middle sums we have rearranged the terms according to increasing powers of $x^n$. Consequently the conditions are $k+l=n$ resp. $100k+l=n$ besides $k,l\geq 0$.


*In the right-hand sums we eliminate $l$ by substituting $l=n-k$ resp. $l=n-100k$. Consequently is the upper limit of the inner sum $n$ resp. $\left\lfloor\frac{n}{100}\right\rfloor$.
