So I know there is a well-known straightforward way to expand something like


and that there are formulas which allow us to expand trinomials and multinomials in general. My question is,

Is there any known way to expand something like $$\left[\sum_{k=0}^{\infty} a_k\right]^n$$ or at least to determine the first few terms?

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    $\begingroup$ One needs to be a bit careful here: If the original series only converges conditionally, then the ordering the of the terms is important. $\endgroup$ Commented May 12, 2016 at 12:52
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    $\begingroup$ You may find this question of interest $\endgroup$
    – MPW
    Commented May 12, 2016 at 12:52
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    $\begingroup$ @Travis: Of course, power series are absolutely convergent on the interior of their disk of convergence, so this would only be a problem on the "edge", if OP knew in advance he was dealing with a power series, for example. +1 for your observation. $\endgroup$
    – MPW
    Commented May 12, 2016 at 12:55
  • $\begingroup$ @MPW: Yes, and your remark is particularly useful here in that power series are surely the place where powers of series arise the most. In that setting, of course there's a preferred ordering for the terms, too, given by the degree. (+1) $\endgroup$ Commented May 12, 2016 at 13:11

2 Answers 2


Here is my attempt at a solution, Let $f(x) = \sum_{k=0}^{\infty}a_k x^k$, then we have that \begin{align*} f(x)^n &= \left( \sum_{k=0}^{\infty}a_k x^k \right)^n \\ &= \sum_{k=0}^{\infty} \left(\sum_{\substack{0 \leq r_1,\ldots,r_n \leq k \\ r_1 + \cdots + r_n = k}}a_{r_1} \cdots a_{r_n}\right) x^k \\ \end{align*} Now for any specific choice of $r_1,\ldots,r_n$ satisfying the condition $$0 \leq r_1,\ldots, r_n \leq k,\quad r_1 + \cdots + r_n = k$$ there are, say, $m \leq n$ many distinct elements in the set $S = \{r_1,\cdots,r_n\}$, call these distinct elements $$s_1,\ldots,s_m$$ and define $$N(s_i) = \# \text{ of times $s_i$ appears in $(r_1,\ldots,r_n)$}$$ then we see that the number of times $$a_{r_1}\cdots a_{r_n}$$ is counted is equal to $$C(r_1,\ldots,r_n) := \frac{n!}{N(s_1)! \cdots N(s_m)!} \quad \text{(The multinomial coefficient)}.$$ Thus we have $$f(x)^n = \sum_{k = 0}^{\infty} \left( \sum_{\substack{0 \leq r_1 \leq\ldots \leq r_n \leq k \\ r_1 + \cdots + r_n = k}}C(r_1,\ldots,r_n)a_{r_1} \cdots a_{r_n} \right) x^k$$ Notice that now the inner sum is over a smaller set and hence (in theory) makes the calculation easier.


If you know multinomials, then the result is the same: the series will contain infinite term, but each term will be composed by at maximum $n$ different $a_k$, and you can determine the coefficient of $a_{i_1}^{r_1}\dots a_{i_s}^{r_s}$, with $r_1+\dots+r_s = n$ by the multinomial $$ \frac{n!}{r_1!\dots r_s!} $$


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