Re: infinite product

In article <6334718.1143190534402.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxx> wrote:

Could you please help me to calculate the following product
\prod_{n=2}^{\infty} (n^2-1)/(n^2+1).

In the case \prod (n^3-1)/(n^3+1) we can easily cancel out the
multipliers in the numerator and denomonator and it can easily be
proved that the values in this case is 2/3. But i have no ideas to
deal with our case with squares.

In one of those twists of fate that make one wonder whether the
universe can really be rational and impersonal, this question shows
up just half a day after I had picked up the December 1999 issue
of the American Mathematical Monthly, whose second page begins
with just this infinite product (in an article by Borwein and Corless,
"Emerging Tools for Experimantal Mathematicians"). They give the value
pi/sinh(pi) .

They also give two related infinite products:

K3 = \prod (1 + 1/k + 1/k^2)^2/(1 + 2/k + 3/k^2)

K4 = \prod (1 + 1/k + 1/k^2 + 1/k^3)^2/(1 + 2/k + 3/k^2 + 4/k^3)

whose numerical values are approximately 1.8489 and 1.7974 respectively.
They give a formula for K3 but state that they don't know one for K4.

The formula for K3 is stated without proof, and indeed, it is not
at all clear that they have a proof of the given formula for K3.
Maple -- the version I use was written after 1999! -- also evaluates K3
but not K4. This raises the question of to what extent one can give
closed-form evaluations of F(P,Q) = \prod P(1/k)/Q(1/k) where P and Q
are polynomials. (The OP's problem asks about F(1-X^2, 1+X^2) .)

It seems to me the products will diverge (possibly diverging to zero)
unless P(0)=Q(0), in which case we can normalize P and Q so that
P(0)=Q(0)=1, as in the examples shown here.

Clearly F(P,Q) = F(P,R)/F(Q,R) as long as all three products converge,
which will happen as long as P'(0)=Q'(0)=R'(0) ; so it will suffice
for our purposes to handle cases with R(x) = 1 + a x, where a = P'(0).
The product is then determined by P alone so let us write
G( P ) = F( P, 1 + P'(0) x ). Note that if P(x) = 1 + a x, then G(P)=1.

Finally note that G( P1 P2 ) = G( P1 ) G( P2 ) F( (1+ax)(1+bx), 1+(a+b) x ),
so using the Fundamental Theorem of Algebra we can reduce this to the
evaluation of G( (1+ax)(1+bx) ) , which can be written e.g. as a product
of values of the Gamma function.

So apparently all products of the same general forms as the OP's product
and the K2 and K3 above permit an expression in closed form.

dave


.

Relevant Pages

  • Re: infinite product
    ... the numerator and denomonator and it can easily be proved that the values in ... Mathematica anyway. ... which seems to correspond to the numerical solution. ...
    (sci.math)
  • Re: infinite product
    ... the numerator and denomonator and it can easily be proved that the values in ... Mathematica anyway. ...
    (sci.math)
Quantcast