Vtyjkyuk

MathWorld (visit http://mathworld.wolfram.com/PiFormulas.html) states that

$displaystyle lim_jtoinftyprod_n=j^2jfracpi2tan^-1n=4^frac1pi$
(equation 130)

With a calculator, I find a general formula:

$displaystyle lim_jtoinftyprod_n=j^mjfracpi2tan^-1(kn)=m^frac2kpi$

But I have not found a proof for this. Any proof would be appreciated.

It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write

$$ P_j = P_j(m,k) = prod_n=j^mj fracpi2arctan(kn) $$

for the expression inside the limit. Then using the relation $arctan(x) = fracpi2 - arctan(frac1x)$ which holds for $x > 0$, we find that

$$ logleft(fracpi2arctan(x)right)
= -logleft(1 - tfrac2piarctanleft(tfrac1xright) right)
= tfrac2pi x + mathcalOleft(x^-2right)
quad textas quad x to infty.$$

So it follows that

$$
log P_j
= sum_n=j^mj left( logfracpi2 - logarctan(kn)right)
= sum_n=j^mj left( frac2pi kn + mathcalO(n^-2) right).
$$

By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_1^m frac2pi kx , dx
= frac2pi klog m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^frac2pi k $ as expected.