15
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I have crafted an infinite product that converges to Zeta[2] very slowly!

  Product[(1296 n^4 (1 + (1 + n)^3)) / 
    ((-1 + 36 n^2)^2 (-1 + (1 +  n)^3)),{n, 1, 50000}]

I am without a computer for 2 more days and have been using Wolfram|Alpha, but it doesn't want to go much beyond the above limit. I would like someone to verify the function to make sure I haven't missed anything.

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  • 2
    $\begingroup$ @ Fred Kline: how did you derive your interesting formula essentially different from the Euler product? $\endgroup$ – Dr. Wolfgang Hintze Sep 7 '15 at 10:58
  • $\begingroup$ @Dr.WolfgangHintze, A few years ago I found a product that produced $\frac{\pi}{3}$ using multiples of $6.$ Yesterday I decided to square it and insert something to multiply by $\frac{3}{2}$. It seems to work. $\endgroup$ – Fred Kline Sep 7 '15 at 12:08
  • $\begingroup$ @Dr.WolfgangHintze, It's the Euler product using multiples of 6. $\endgroup$ – Fred Kline Sep 7 '15 at 12:27
  • $\begingroup$ @ Fred Kline: as to the name "Euler product" please see my answer $\endgroup$ – Dr. Wolfgang Hintze Sep 7 '15 at 14:40
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Product[
  (1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3))
  , {n, 1, ∞}
  ] === Zeta[2]
True
Zeta[2]
π^2/6
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  • 2
    $\begingroup$ Error decreases approximately as $\frac{1}{10n}$. $\endgroup$ – rhermans Sep 7 '15 at 10:10
12
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Amplifying on answer by @rhermans

f[m_] = Product[(1296 n^4 (1 + (1 + n)^3))/((-1 + 
        36 n^2)^2 (-1 + (1 + n)^3)), {n, 1, m}]

(* (Pi^2*Gamma[1 + m]^3*Gamma[3 + m])/
   (6*(3 + 3*m + m^2)*Gamma[5/6 + m]^
        2*Gamma[7/6 + m]^2) *)

This product converges

Limit[f[m + 1]/f[m], m -> Infinity]

(* 1 *)

Limit[f[m], m -> Infinity]

(* Pi^2/6 *)

Product[(1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3)), {n, 1, 
  Infinity}]

(* Pi^2/6 *)

% === Zeta[2]

(* True *)

LogLinearPlot[{f[m], Zeta[2]}, {m, 1, 100},
 Epilog -> 
  Inset[LogLinearPlot[{f[m], Zeta[2]}, {m, 75, 100000}], {Log[25], 1.55}]]

enter image description here

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  • 1
    $\begingroup$ +1 for nice plot. I've learned something new. $\endgroup$ – Fred Kline Sep 7 '15 at 12:22
8
+100
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I tried to find an even simpler product. Here's my solution:

$$ \zeta(2) =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)}$$

In Mathematica

Product[ 1/(1 - 1/(4 n^2)) 1/(1 - 1/(36 n^2)), {n, 1, \[Infinity]}]

(* Out[76]= \[Pi]^2/6 *)

We can derive this from the well-known product formula of the sine

Product[1 - x^2/n^2, {n, 1, \[Infinity]}]

(*
Out[85]= Sin[\[Pi] x]/(\[Pi] x) 
*)

considering that

\[Pi] x/Sin[\[Pi] x] /. x -> 1/2

(*
Out[79]= \[Pi]/2
*)

and

\[Pi] x/Sin[\[Pi] x] /. x -> 1/6

(*
Out[80]= \[Pi]/3 
*)

The product of these two expressions give [Pi]^2/6. Finally, it is easy to transform the corresponding infinite products into the form provided above.

Remark: Although Euler used the product formula for the sine in his famous proof that the infinite sum of the inverse squares is equal to [Pi]^2/6, normally an Euler product is a product over primes, such as the one defining the zeta function:

$$\zeta(s) =\prod _{n=1}^{\infty } \frac{1}{1-p_n^{-s}}$$

EDIT #1

It is not difficult to prove that for any positive integer m we can write $$\zeta (2 m)=\prod _{n=1}^{\infty } a (n)$$

Where the $a(n)$ are rational functions of n.

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This is going to be an alternative answer to Dr. Wolfgang Hintze's question. Consider a limit: \begin{equation} g := \prod\limits_{n=1}^\infty \frac{1}{\left(1-\frac{A^2}{n^2}\right)\left(1-\frac{B^2}{n^2}\right)\left(1-\frac{C^2}{n^2}\right)\left(1-\frac{D^2}{n^2}\right)} \end{equation} Taking logs we have: \begin{equation} \log(g) = - \sum\limits_{n=1}^\infty \left[\log(1-\frac{A^2}{n^2}) + \log(1-\frac{B^2}{n^2})+\log(1-\frac{C^2}{n^2})+\log(1-\frac{D^2}{n^2})\right] = \sum\limits_{l=1}^\infty \frac{1}{l} (A^{2 l} + B^{2 l}+C^{2 l}+D^{2 l}) \zeta(2 l) \end{equation} Now from Faulhaber's formula we know values of the zeta function at positive integers (see for example Wikipedia page on zeta function). Therefore we have: \begin{eqnarray} \log(g) =\\ (-1) \left\{ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi A)^{2 l}}{(2 l)!}\frac{1}{2 l} + \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi B)^{2 l}}{(2 l)!}\frac{1}{2 l}+ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi C)^{2 l}}{(2 l)!}\frac{1}{2 l}+ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi D)^{2 l}}{(2 l)!}\frac{1}{2 l} \right\} = \\ \sum\limits_{t=2\pi\left\{A,B,C,D\right\}} \log(\Gamma(1+\frac{t}{2 \pi})\Gamma(1-\frac{t}{2 \pi})) \end{eqnarray} Therefore the limit reads: \begin{equation} g = \prod\limits_{t=\left\{A,B,C,D\right\}} \Gamma(1+t) \Gamma(1-t) = \pi^4 \frac{A B C D}{\sin(\pi A)\sin(\pi B)\sin(\pi C)\sin(\pi D)} \end{equation}

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This is not an answer. I have additional observations about the answer by Dr. Wolfgang Hintze.
In response to his first comment: From here, when we omit the first two primes from Euler's product, we get a square: $$\frac{\pi^2}{9} =\prod _{n=3}^{\infty } \frac{1}{1-p_n^{-2}},$$ then substituting $6n$ for $p_n,$ we get the square root: $$\frac{\pi}{3} =\prod _{n=1}^{\infty } \frac{1}{1-(6n)^{-2}}.$$

Now about his simplified product: When we omit the left-hand factor of the denominator, we get $\frac{\pi}{3}$ as above.

The left-hand factor of the denominator simplifies my machinations. Very nice.

However, that denominator has another interesting pattern: two odd primes bracket the $4$ as $(3,5)$ and all other odd primes intermittently bracket multiples of $6$ as $(5,7),(11,13),\dots$ So we have a product of all the primes which equals the product of all the points where odd primes may be found:

$$\zeta(2) =\prod _{n=1}^{\infty } \frac{1}{1-p_n^{-2}} =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)}$$

Multiples of $n$ where $0\equiv n \mod 3$ cause the left-hand denominator to collide with a previous right-hand denominator Because we omitted the prime $2,$ we will have a missing ratio: $\frac{1}{2}$ or $\frac{2}{1}.$

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  • $\begingroup$ @ Fred Kline: Honestly, I don't understand your remark on the "interesting pattern". BTW you should write 2 instead of s in the Euler products. $\endgroup$ – Dr. Wolfgang Hintze Sep 8 '15 at 12:21
  • $\begingroup$ @Dr.WolfgangHintze, Thanks for the heads-up re: $s.$ The interesting pattern is that the prime product uses primes only. We use multiples of $4,6$ which are the mid-points where odd primes can occur. As soon as I get my activation code for Mathematica later today, I plan to look at both functions to see the differences/commonalities. It might not be special? $\endgroup$ – Fred Kline Sep 8 '15 at 13:02
  • 1
    $\begingroup$ @ Fred Kline: you might wish to give the proof of the theorem I presented in my EDIT #1. Try to apply it to zeta(4), for instance, before generalizing too much from one case. $\endgroup$ – Dr. Wolfgang Hintze Sep 8 '15 at 13:14

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