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We know that $$\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6},$$ but what about the product of the reciprocal of the squares: $$\prod_{m=1}^\infty \frac{1}{m^2}?$$ Do we use a different product representation to compute this? Maybe the cosine product $$\cos{x} = \prod_{m=1}^\infty \left(1-\frac{x^2}{\pi^2\left(m-\frac{1}{2}\right)^2}\right).$$

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    The product rapidly diverges to zero. – Angina Seng May 15 '20 at 16:36
  • Look at the sequence of partial products: $1,\frac14,\frac1{36},\ldots$ – user170231 May 15 '20 at 16:36
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    @Chickenmancer I meant what I said. – Angina Seng May 15 '20 at 16:40
  • Products converge if and only if it goes to some non-zero number. Zero is counted as divergence. – QC_QAOA May 15 '20 at 16:45
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    For infinite products $\to 0$ is most often called "diverging to zero". : https://en.wikipedia.org/wiki/Infinite_product – Winther May 15 '20 at 16:45
  • My apologies @AnginaSeng. I wasn't aware of the convention. – Chickenmancer May 15 '20 at 16:56
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    @AnginaSeng To be fair, diverging to zero is a contamination. Either you say it diverges, or you say it converges to $0$. Just like saying that a sum diverges, or it converges to infinity (the difference is wether you allow $0$ or $\infty$ in the set of possible limits). You cannot diverge to something as diverging means to go away from something, and converging means to get closer to something. Some people say it though... – Jens Renders May 15 '20 at 16:58

4 Answers4

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$$\prod_{m=1}^n \frac{1}{m^2}=\frac1{(n!)^2},$$ which tends to $0$ at the speed of light.

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Note that one clearly has

$$0 \le \prod_{m=1}^N\dfrac{1}{m^2} \le \dfrac{1}{N^2}.$$

The RHS tends to $0$ as $N \to \infty$ which gives us that the sequence of partial products converges to $0$. (Which is the same as saying that the product diverges (!) to $0$.)

Thus, to answer your question: Yes, the product does have a closed form!

Aryaman Maithani
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  • Infinite products whose partial products converge to $0$ are considered divergent. – Sam May 15 '20 at 17:06
  • @Sam: Ah, yes, terminology. Fixed! – Aryaman Maithani May 15 '20 at 17:11
  • @Sam Altough it is correct to say that the product diverges, it is also correct to say that the product converges to $0$, and it is also correct to say that the product is equal to $0$. – Jens Renders May 15 '20 at 17:17
  • @JensRenders That's just the accepted terminology for infinite products, look it up in any analysis textbook. – Sam May 15 '20 at 17:33
  • @Sam I just said it was correct... – Jens Renders May 15 '20 at 17:34
  • @JensRenders It's not correct to say that the product converges to $0$ – Sam May 15 '20 at 17:35
  • @Sam The expression is often used and is just shorthand for "the sequence of partial products converges to zero". It is more accurate though, to say that the product is equal to zero. – Jens Renders May 15 '20 at 17:39
  • @JensRenders As I said, just look it up. If we accept "converges to $0$" as a valid expression for infinite products, we would have to add the additional assumption "assume the product is not $0$" to just about every theorem there is about convergent infinite products. For example: If $a_n \ge 0$ for all $n$, then $\Pi(1+a_n)$ converges $\iff \sum a_n$ converges. Or, if $\Pi a_n$ converges and $b_n = a_n$ for all but finitely many $n$ then $\Pi b_n$ also converges. Both of these results are false if you allow "converges to $0$" as valid. – Sam May 15 '20 at 17:57
  • @Sam I don't have to look anything up as I am well aware of the intricacies. I am also aware why many authors call it divergence instead of calling it convergence to $0$. Still many other authors do exactly what you say in your last comment: always say convergence to a non-zero real number, or simply convergence in $\mathbb{R}^+ = ]0,\infty[$. – Jens Renders May 15 '20 at 18:01
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For a sum to converge, it is necessary that the terms converge to $0$ (the neutral element for the sum). $\frac{1}{n^2}$ satisfies this.

For a product to converge to a nonzero value, it is necessary that the factors converge to $1$ (the neutral element for the product). $\frac{1}{n^2}$ does not satisfies this, so the product diverges. In this case, the product is $0$.

But because $\frac{1}{n^2} \to 0$, we do of course have that $$1+\frac{1}{n^2} \to 1$$ and $$1-\frac{1}{n^2} \to 1$$

So perhaps more interesting product analogies of the sum you mention are $$\begin{aligned} \prod_{n=1}^\infty 1+\frac{1}{n^2} &=\frac{\sinh(\pi)}{\pi} = \frac{-e^{-\pi}}{2\pi} + \frac{e^\pi}{2\pi}\\ \prod_{n=2}^\infty 1-\frac{1}{n^2} &= \frac{1}{2} \end{aligned} $$

Jens Renders
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Assuming you can use the fact that $n! \to \infty$ it follows that $\frac{1}{n!} \to_n 0$ and since $\frac{1}{n!^2} \leq \frac{1}{n!}$ it converges to 0 by squeeze lemma

Alex
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