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In the conditionally convergent infinite product $\lim_{k\rightarrow \infty}{\prod'}_{n=-k}^k{(1-\frac{z}{n\pi})}{e^{\frac{z}{n\pi}}}$ (where $\prod'$ indicates n=0 is excluded), we can insert infinitely many $e^{-\frac{z}{n\pi}}e^{\frac{z}{n\pi}}, n\in\mathbb{Z}-\{0\}$ into the series (and then move around some of $e^{\frac{z}{m\pi}},m\in\mathbb{Z}-\{0\}$) without changing the limit of the product (but change the infinite product to an absolutely convergent one.

In general,

Question 1. Can we insert any finite product $a_1a_2\dots a_n$, which equals unity, into an infinite product? (It seems yes.)

Can we insert any infinite product $a_1a_2\dots a_n\dots$, which equals unity, into an infinite product?

(By saying this I imply inserting an infinite product $X$ converging
to unity won't change the limit of the infinite product (that we
insert $X$ into).)

Does similar conclusion hold for infinite series?

Question 2. Also, while commutative law is usually true only for absolutely convergent infinite product/series, it seems associative law is always valid for any (absolutely and conditionally) convergent infinite product/series, right?

(by saying that association law holds,I imply that after we re-bracketing items (redefining each item, see example $\lambda_n, > \eta_n$ below),

(1). the limit won't change, and

(2). a conditionally convergent infinite series/product is still conditionally convergent one;

and the same for an absolutely convergent infinite series/product.

Well, I guess very likely it will not change the limit,(possibly it and its reason similar to what's discussed in my another post Why does an infinite series that conditionally converges, when its items rearranged, tend to a different limit?? And it seems surer when we consider infinite product can be written as infinite series of log) though that's not strictly true for 'abs conv', e.g. de-bracketing $[(1-\frac{z}{\pi})(1+\frac{z}{\pi})][(1-\frac{z}{2\pi})(1+\frac{z}{2\pi})]\dots$ will make it not 'abs conv'.)


PS:the above 'insertion' is sort of associative law, I guess.

And commutative law actually implies that if all items/'bracketed/associated items' $a_n$, whose $\sum \log a_n$ converges absolutely, can be moved around freely in the sequence.

Example 1. If $\lambda_1\lambda_2\lambda_3\lambda_4\lambda_5\lambda_6\lambda_7\lambda_8\lambda_9\dots=\lambda_1(\lambda_2\lambda_3\lambda_4)\lambda_6(\lambda_7\lambda_8)\lambda_9\dots=\eta_1\eta_2\eta_3\eta_4\eta_5\dots,$ and $\sum \log \eta_n$ converges absolutely, then $\eta_1\eta_2\eta_3\eta_4\eta_5\dots$ converges absolutely, even if $\lambda_1\lambda_2\lambda_3\lambda_4\lambda_5\lambda_6\lambda_7\lambda_8\lambda_9\dots$ doesn't.

Example 2. More specifically, $(1-\frac{z}{\pi})(1+\frac{z}{\pi})(1-\frac{z}{2\pi})(1+\frac{z}{2\pi})\dots$ converges conditionally, while $[(1-\frac{z}{\pi})(1+\frac{z}{\pi})][(1-\frac{z}{2\pi})(1+\frac{z}{2\pi})]\dots$ ([...] is regarded as ONE item) converges absolutely. Therefore, absolute convergence and commutative law are strongly linked with how we bracket items/what we regard as ONE item.


Edited to add: I post a rough answer below.

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  • $\begingroup$ What do you mean precisely by does an “association law” hold? The commutative property for absolutely convergent series means precisely that permuting the product (even infinitely) will result in the same value (finite permutations world in the conditional case). And what do you mean by “insert an infinite product”? $\endgroup$ Commented Aug 8, 2020 at 16:40

3 Answers 3

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An infinite product $\prod_{k=1}^\infty a_k$ converges to $L \neq 0$ iff the sequence of partial products $b_n := \prod_{k=1}^n a_k$ has limit $L$.

From this definition it follows that if we “insert” any nonzero finite product at any point in the sequence — or even interspersed throughout the sequence — the new sequence converges to the limit of the old sequence times the value of the finite product. Let’s say we insert a finite number $j$ of new elements, the last of which is at index $i$, to get a new sequence $\{a’_k\}$ with partial products $\{b’_k\}$. Then we know $b’_{k+j} = C b_k$ for all $k \geq i$ where $C$ is the value of the product we’ve inserted; this suffices to show that the limit $\lim_{k \to \infty} b’_k = CL$.

One way to interpret the “associativity” question is to say: Let’s say we partition the natural numbers into finite, contiguous “runs” $\{I_k\}$ (e.g., runs of size two: $I_k := \{2k,2k+1\}$) and condense the infinite product of the sequence into the (infinite) product of the (finite) products over each run $$\prod_{k=1}^\infty \prod I_k. $$ A similar partial product argument should show that these two products always converge to the same limit.

If two infinite products of sequences $\{a_k,a’_k\}$ each converge and we combine them into one sequence $a_1,a’_1,a_2,a’_2,\ldots$ via alternation, another similar argument should show that the infinite product of the new sequences converges to the product of the old sequences via partial products and this. More complicated methods of interspersion should also work (e.g., inserting finite, contiguous runs from $\{a’_k\}$ at various points), as long as order is preserved and you can match partial products of the new sequence with pairs of products of the old sequence.

The moral is that finite operations (permutations, insertions, etc.) are generally okay because they only have a “substantive” impact on a finite number of partial products, so we don’t need to worry about convergence issues; some infinite operations can also be okay, for instance, in cases where the partial products can be matched in an order-preserving way (e.g., splitting the sequence into pairs and swapping each pair).

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  • $\begingroup$ it looks so, but since we can't swop items $a_n$, is it really okay that we add items? $\endgroup$ Commented Aug 8, 2020 at 16:41
  • $\begingroup$ Would you like to provide any related textbook (for more details) with formal proof showing strictly why we can do so? $\endgroup$ Commented Aug 9, 2020 at 0:42
  • $\begingroup$ I'm not aware of any particular sources for these statements. I am happy to edit the above to make any of the proofs more formal. $\endgroup$ Commented Aug 9, 2020 at 0:45
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This is not a strict answer, and contains some guessing.

I guess I finally realize something, by considering the actual origin and practical application of infinite series or product.

An infinite series or product, e.g. 1-1/2+1/3-..., $\frac{1}{2}\frac{3}{4}\dots$, is just a 'long' algebraic expression, (similar to infinite fraction, etc.)--in the same sense that finite sum and product 1-1/2+1/3, $\frac{1}{2}\frac{3}{4}\frac{5}{6}$ are algebraic expression--and all the laws for these algebraic operations of certain algebraic structure hold, as long as we just do finite operations on infinite series or product, and similarly perhaps we can insert finitely many items, together or separately, whose sum or product equals unity in the algebraic structure. Things go uncertain only when we need to do infinitely many such operations.

  1. Association law seems to hold even if we use it infinitely many times (I don't know why). As a result of this, we need no bracket or even define 'item' for an infinite series/product.

  2. We need to do so when we consider absolute convergence series/product, for then we need to add | |. It's not normal bracket--and so association law naturally doesn't hold for absolute convergence of infinite series, it doesn't hold even for finite sum, e.g. |x+y|$\leq$|x|+|y|; it actually greatly change the series/product, it's calculation/function itself, it does produce a new series/product (absolute convergence means the new series/product is convergent, and it is related to but not exactly about the original series/product, for we not only do | | operation, but define 'items' in a way that can be quite arbitrary, e.g. |1|+|-1/2|+|1/3-1/4+1/5|+...=1+1/2+7/60+... which looks completely different from the original series), not just about changing order of calculation.

  3. In multiplication of finite sum, we can arbitrarily bracket items, and then with distribution law we get the same result whatever way we do bracketing, so there we need no bracket. Consistent with what's stated above, if we just need to use distribution law finitely many times, then the laws should be valid for infinite series as well and we need no bracket. But if we need to use the law infinitely many times, perhaps we should define 'items' clearly, so we need bracket here.

3-1. Further more, if with the multiplication we want not double, triple,... infinite series, but a usual infinite series (say $P$), then we need to cluster the new items (say $p_{m,n}$) from multiplication into pieces (e.g. we cluster $\{p_{n-k,k}\}=\{a_nb_1, a_{n-1}b_2,\dots, a_1b_n\}$), if so, for that purpose we surely need bracket, otherwise we don't have definite 'items' $p_{m,n}$, not to say proper cluster of them.

3-2. (We need 'clusters' not really because we have to define what are items for $P$, but just for we need to define order of summation for $$p_{m,n}$, e.g. with above ways of clustering, we get

$$P=a_1b_1+a_2b_1+a_1b_2+a_3b_1+ a_{3-1}b_2+\dots+a_1b_3+a_4b_1+a_{4-1}b_2+\dots+ a_1b_4\dots,$$

the clustering enables us to define order of summation this way, (or equivalently another order where any two elements in $\{a_nb_1, a_{n-1}b_2,\dots, a_1b_n\}$ swops freely, as long as total number of 'swops' are finite). But actually we can define items to be arbitrarily nearby strings like $(a_1b_2+a_3b_1+ a_{3-1}b_2)$, not to be 'clusters', and that doesn't at all affect the convergence (not considering absoluteness) and limit of the series.)

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We can prove something way more general: any order-preserving interlacing of two convergent infinite products converges, and its value is equal to the product of the individual values.

Suppose $\{ a_k \}_{k \geq 1}$ is a sequence of positive real numbers so that $\prod_{k} a_k$ converges to the value $A$, and similarly, let $\{ b_k \}_{k \geq 1}$ be a sequence of positive real numbers so that $\prod_{k} b_k$ converges to the value $B$. (To get the situation in the original post, suppose that $A = 1$; for the special case of finite products, suppose that $A = 1$ and $a_k = 1$ for all sufficiently large $k$.)

We recall here the definition of convergence of an infinite product: $\prod_{k} a_k$ converges iff the sequence of partial products $A_n = \prod_{k=1}^n a_k$ converges: $\lim_{n \to \infty} A_n = A$. We use the corresponding notation for $B_n = \prod_{k=1}^n b_k$ and make the same assumption that $\lim_{n \to \infty} B_n = B$.

Let $\alpha, \beta: \Bbb{N} \rightarrow \Bbb{N}$ be any two order-preserving injections so that $\alpha(\Bbb{N}) \cap \beta(\Bbb{N}) = \emptyset$ (there's no overlap between their images) and $\alpha(\Bbb{N}) \cup \beta(\Bbb{N}) = \Bbb{N}$ (every natural number is in one set or the other set). Define the interlaced sequence $c_k$ to be $c_k = a_{\alpha^{-1}(k)}$ if $k \in \alpha(\Bbb{N})$, and $c_k = b_{\beta^{-1}(k)}$ if $k \in \beta(\Bbb{N})$.

Theorem: $\prod_{k} c_k$ converges and equals $AB$.

Proof: Consider the sequence of partial products $C_n = \prod_{k=1}^n c_k$. Let $M_\alpha(n) = \max \alpha(\Bbb{N}) \cap \{ 1, 2, ..., n \}$, and similarly, $M_\beta(n) = \max \beta(\Bbb{N}) \cap \{ 1, 2, ..., n \}$. For sufficiently large $n$, these intersections will be nonempty, and so $M_\alpha(n), M_\beta(n)$ will be well defined, and $\lim_{n \to \infty} M_\alpha(n) = \lim_{n \to \infty} M_\beta(n) = \infty$. Then for $n$ sufficiently large, $$C_n = A_{M_\alpha(n)} B_{M_\beta(n)},$$ so $$\lim_{n \to \infty} C_n = \lim_{n \to \infty} A_{M_\alpha(n)} \lim_{n \to \infty} B_{M_\beta(n)} = AB.$$

Re-associating the factors doesn't change the values of the partial products, but switching the orders of the factors can change the values of the partial products, which can affect the limiting value, or even whether or not there is convergence. This is why regrouping (without changing order) doesn't affect convergence/limiting value but rearranging (i.e. changing order) does, unless you have absolute convergence.

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