Evaluate these infinite products $\prod_{n\geq 2}(1-\frac{1}{n^3})$ and $\prod_{n\geq 1}(1+\frac{1}{n^3})$

Question

What is

$\prod\limits_{n\geq 2}(1-\frac{1}{n^3})=?$
$\prod\limits_{n\geq 1}(1+\frac{1}{n^3})=?$

I am sure about their convergence. But don't know about exact values. Know some bounds as well. For example first one is in interval (2/3,1) and second one is in (2,3).

What does this have to do with zeta-functions? Those are generally products over primes, not over all $n$. — Gerry Myerson, Feb 17 '14 at 11:05
@Gerry Myerson : may be something to do with multi zeta values?? not sure.. — user41481, Feb 17 '14 at 11:07
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. — Did, Feb 17 '14 at 11:08
If you find a connection to zeta-functions, you can edit them back into the tags. — Gerry Myerson, Feb 17 '14 at 11:10
@Did : I am sure about their convergence. But don't know about exact values. Know some bounds as well. For example first one is in interval (2/3,1) and second one is in (2,3). — user41481, Feb 17 '14 at 11:15
According to Maple the second value is $$\frac{\cosh(\frac{1}{2}\pi \sqrt{3})}{\pi} \approx 2.4282\dots$$ and the first is one third of the second. — gammatester, Feb 17 '14 at 11:33
@gammatester : Will you please explain how did you get that expression involving cosh? — user41481, Feb 17 '14 at 12:42
With the Maple command b:=product(1+1/n^3, n=1..infinity); you get a first result $$b = \frac{\sin(\pi(\frac{1}{2} + \frac{1}{2} i \sqrt{3}))}{\pi}\cdot$$ And because the result is real a second command Re(b) gives the expression for the second product: $$\frac{\cosh(\frac{1}{2} \pi \sqrt{3}))}{\pi}$$ Note that Wolfram Alpha gives the $\cosh$ result in one step. — gammatester, Feb 17 '14 at 12:57
@Biswa cf. http://mathworld.wolfram.com/InfiniteProduct.html formulae (20), (22) and (30). There is also a reference to a general formula. — boxdot, Feb 17 '14 at 13:04
The first question was asked (and answered) here: http://math.stackexchange.com/questions/319784/evaluating-prod-limits-n-2-infty-left1-frac1n3-right — Martin Sleziak, Feb 17 '14 at 15:05

score 8 · Answer 1 · edited Apr 13 '17 at 12:20

Ramanujan has already answered these two questions $100$ years ago!
The same questions have already been answered here, by Mhenni Benghorbal, and here, by Claude Leibovici.
Generally speaking, $\quad\displaystyle\prod_{n=1}^\infty\bigg(1\pm\frac{x^{2k+1}}{n^{2k+1}}\bigg)=\pm\frac1{x^{2k+1}\cdot\displaystyle\prod_{j=1}^{2k+1}\Gamma\bigg((\pm1)^{^{\frac j{2k+1}}}x\bigg)}$ . When $x=1$ and the sign is negative, in order to find the value of your product, you must obviously divide through $1-\bigg(\dfrac xn\bigg)^{2k+1}$ But since both the numerator and denominator are $0$, we must use l'Hopital. For $k=1\iff2k+1=3$, this will lead you to the above-linked results, which have also been confirmed by Gamma Tester in this post's comment section.
As far as the relationship to the $\zeta(3)$ function is concerned, this is clear from Euler's approach to the Basel problem, which connects $\zeta(2)$ with the infinite product for $\dfrac{\sin x}x$ , as well as from the fact that, in general, the convergence of $\prod(1+a_k)$ is the same as that of $\sum a_k$. Hope this helps!

score 2 · Answer 2 · answered Feb 17 '14 at 12:02

Just one quick observation.

Let, $P_1$ denotes the first product and $P_2$ the second. Then, since $P_1$ converges and is positive, $$\log{P_1}=\sum_{n\ge 2}\log\left(1-\frac 1{n^3}\right)=-\sum_{n\ge 2}\sum_{k\ge 1}\frac{1}{kn^{3k}}=-\sum_{k\ge 1}\frac{\zeta(3k)-1}{k}$$

Similarly, $$\log P_2=\sum_{k\ge 1}\frac{(-1)^{k-1}\zeta(3k)}{k}$$

Evaluate these infinite products $\prod_{n\geq 2}(1-\frac{1}{n^3})$ and $\prod_{n\geq 1}(1+\frac{1}{n^3})$

2 Answers2

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