Can we formally multiply out infinite products?

Question

I came across Euler's proof that $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$. One of the ingredients of the proof uses $$\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)=1-\frac{x^2}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}+\text{higher order powers}$$ i.e. it seems that we can treat an infinite product as a finite product in the sense that we can "multiply it out and collect like powers". How can this be justified? If it cannot be justified, can you give a counterexample?

Assume that the infinite product converges absolutely.

Answer 1

The fact that many infinite products in Euler have some properties of finite products is explained in this publication in Journal for General Philosophy of Science. Briefly, Euler's procedures can be formalized in a hyperreal extension of $\mathbb R$ where unlimited integers (greater than every standard integer) are available. Infinite products are often approximated infinitely well by hyperfinite products. The latter have all the first-order properties of ordinary finite products. The article linked above analyzes in detail Euler's proof of the infinite product decomposition for the sine function.

Answer 2

Suppose that you write instead $$\prod_{n=1}^\infty \left(1-\frac{x^2}{a_n^2\,\pi^2}\right)$$ Expand it as a Taylor series to obtain $$1-\frac {x^2}{\pi^2}\sum_{n=1}^\infty \frac 1{a_n^2}+O(x^4)$$ Just let $a_n=n$.

The next term to $O(x^8)$ would be $$+\frac {x^4}{\pi^4}\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac 1{a_m^2\, a_n^2}$$ Now, is strats to be more difficult.

Answer 3

$\newcommand{\abs}[1]{\left\lvert#1\right\rvert}$ [This is something I wrote in 2005. I've translated it from $\LaTeX{}$ to MathJax, and offer it for what it's worth. I would have to revise quite a lot of stuff to make sure I still understand it fully myself, but it's probably valid. I never finished working on it, because I intended to go on to use Möbius inversion to develop a theory that would generalise the relation between the terms and partial sums of an infinite series, but I never got round to it.]

[In 2015, I came across a relevant Maths.SE answer, which for all I know may be more helpful than my own: https://math.stackexchange.com/a/509318/213690.]

Lemma A. If the unordered sum $\sum_{x \in X} f(x)$ exists and equals $\alpha$, then given any positive $\varepsilon$, there is a finite subset $E$ of $X$ such that \begin{equation}\label{eqn:inf_sub}\tag{1} \abs{ \sum_{x \in Y} f(x) - \alpha } < \varepsilon \end{equation} whenever $Y$ is a subset of $X$ and $Y \supset E$. [Footnote: I think this could also be expressed by saying that the function $Y \mapsto \sum_{x \in Y} f(x)$ is continuous on the directed set $\mathcal{P}(X)$ of all subsets of $X$.]

Proof. By the definition of unordered sum, there exists finite $E \subset X$ such that \begin{equation}\label{eqn:fin_sub}\tag{2} \abs{ \sum_{x \in F} f(x) - \alpha } < \frac{\varepsilon}{2} \end{equation} whenever $F$ is finite and $E \subset F \subset X$.

Suppose now $E \subset Y \subset X$, $Y$ not necessarily finite. $\sum_{x \in Y} f(x)$ is the limit of $\sum_{x \in F} f(x)$ for finite $F$ contained between $E$ and $Y$, so there exists finite $G$ such that $E \subset G \subset Y$ and, if $F$ is finite and $G \subset F \subset Y$, then \begin{equation}\label{eqn:fin_inf}\tag{3} \abs{ \sum_{x \in F} f(x) - \sum_{x \in Y} f(x) } < \frac{\varepsilon}{2}. \end{equation} Taking \eqref{eqn:fin_sub} and \eqref{eqn:fin_inf} together (for an arbitrary choice of such $F$, for instance $F = G$) gives \eqref{eqn:inf_sub}. $\ \square$

Lemma B. Suppose that the unordered sum $\sum_{x \in X} f(x)$ exists and equals $\alpha$. Suppose also that $Y_1, Y_2, \dotsc$ is a sequence of subsets of $X$ such that $$ Y_1 \subset Y_2 \subset Y_3 \subset \dotsb, \quad \bigcup_{n = 1}^\infty Y_n = X. $$ Then \begin{equation}\label{eqn:lim_sum}\tag{4} \lim_{n \to \infty} \sum_{x \in Y_n} f(x) \end{equation} exists and equals $\alpha$. In particular, if $Z_1, Z_2, \dotsc$ satisfies the same properties as the $Y_n$, then \begin{equation}\label{eqn:lim_sum_eql}\tag{5} \lim_{n \to \infty} \left( \sum_{x \in Y_n} f(x) \right) = \lim_{n \to \infty} \left( \sum_{x \in Z_n} f(x) \right). \end{equation}

Proof. The same as Beardon, Theorem 5.2.2, but using his Corollary 5.2.4, and Lemma A above in place of the definition of unordered sum. $\ \square$

Lemma C. Let $(a_{i, j})_{i \geqslant 1, j \geqslant 0}$ be a double sequence of complex numbers such that $a_{i, 0} = 1$ and the series $\sum_{j \geqslant 0} a_{i, j}$ converges absolutely for each $i \geqslant 1$, and the infinite product $\prod_{i \geqslant 1} \sum_{j \geqslant 0} \abs{a_{i, j}}$ converges [Apostol, section 8.26]. Let $\Lambda$ be the countable set consisting of all sequences $\lambda = (\lambda_i)_{i \geqslant 1}$ of non-negative integers almost all equal to 0, and define $d_\lambda = \prod_{i \geqslant 1} a_{i, \lambda_i}$. Then the family $(d_\lambda)_{\lambda \in \Lambda}$ is absolutely summable [Dieudonné, section 5.3], and if $b_i = \sum_{j \geqslant 0} a_{i, j}$ ($i \geqslant 1$), the infinite product $\prod_{i \geqslant 1} b_i$ converges to the unordered sum $\sum_{\lambda \in \Lambda} d_\lambda$ (or diverges to zero if the unordered sum is zero). That is, $$ \prod_{i = 1}^\infty \sum_{j = 0}^\infty a_{i, j} = \sum_{\lambda \in \Lambda} \prod_{i = 1}^\infty a_{i, \lambda_i}. $$

Proof. For $n \geqslant 1$, let $Y_n = \{ \lambda \in \Lambda : \lambda_i = 0 \text{ for all } i > n \}$. Then $Y_1 \subset Y_2 \subset \dotsb$, and $\bigcup_{n = 1}^\infty Y_n = \Lambda$. Also for $m \geqslant 0$, let $E_{mn} = \{ \lambda \in \Lambda : \lambda_i \leqslant m \text{ for } i = 1, \dotsc, n \}$. Then $E_{mn}$ is finite, and $\bigcup_{m = 0}^\infty E_{mn} = Y_n$.

If $A$ is any finite subset of $\Lambda$, then $A \subset E_{mn}$ for some $(m, n)$, and then \begin{equation*} \sum_{\lambda \in A} \abs{d_\lambda} \leqslant \sum_{\lambda \in E_{mn}} \abs{d_\lambda} = \prod_{i = 1}^n \sum_{j = 0}^m \abs{a_{ij}} \leqslant \prod_{i = 1}^n \sum_{j = 0}^\infty \abs{a_{ij}} \leqslant \prod_{i = 1}^\infty \sum_{j = 0}^\infty \abs{a_{ij}} \end{equation*} so the unordered sum $\sum_{\lambda \in \Lambda} d_\lambda$ exists, by Beardon, Theorem 5.2.3(iii).

By Beardon, Theorem 5.2.2 (or the more general Lemma B above), for $n \geqslant 1$, \begin{equation*} \sum_{\lambda \in Y_n} d_\lambda = \lim_{m \to \infty} \sum_{\lambda \in E_{mn}} d_\lambda = \lim_{m \to \infty} \prod_{i = 1}^n \sum_{j = 0}^m a_{ij} = \prod_{i = 1}^n \sum_{j = 0}^\infty a_{ij}. \end{equation*}

Therefore, by Lemma B, \begin{equation*} \sum_{\lambda \in \Lambda} d_\lambda = \lim_{n \to \infty} \sum_{\lambda \in Y_n} d_\lambda = \lim_{n \to \infty} \prod_{i = 1}^n \sum_{j = 0}^\infty a_{ij} = \prod_{i = 1}^\infty \sum_{j = 0}^\infty a_{ij}. \quad \square \end{equation*}

I note in passing that if, for $\lambda, \mu \in \Lambda$, we define $\mu \leqslant \lambda$ iff $\mu_i \leqslant \lambda_i$ for all $i \geqslant 1$, then the set $\Lambda$ itself is directed, each set $\{ \mu : \mu \leqslant \lambda \}$ is finite, and $$ \sum_{\mu \leqslant \lambda} d_\mu = \prod_{i = 1}^\infty \sum_{j = 0}^{\lambda_i} a_{ij} $$ (which despite apperances is a finite expression, on both sides). One could prove a more general version of Beardon, Theorem 5.2.2, in which the sequence $(E_n)$ is replaced by a net (which in this instance happens to be indexed by the set $\Lambda$ itself), so that we would have $$ \sum_{\lambda \in \Lambda} d_\lambda = \lim_{\lambda \to \infty} \sum_{\mu \leqslant \lambda} d_\mu $$ I wouldn't be surprised if this is part of some general theory which I haven't had time to examine. In any case, it probably leads to a more elegant proof using the finite sets $\{ \mu : \mu \leqslant \lambda \}$ instead of the $E_{mn}$ and $Y_n$.

Alternatively (and less elegantly), it might have been possible to make do with the sets $E_{nn}$, but I didn't even seriously consider this, as both of the other possibilities seemed more natural.

References

Tom M. Apostol, Mathematical Analysis, 2nd edition (1974).

Alan F. Beardon, Limits: A New Approach to Real Analysis (1997).

J. Dieudonné, Foundations of Modern Analysis (1969).