Wikipedia:Reference desk/Archives/Mathematics/2013 July 22

= July 22 =

Convergence and divergence of infinite sum of (n^a*sin(n)^b)^-1
Just read from a book that it is not proven whether the case a=3 and b=2 converges or diverges(might be outdated, though), because sometimes the integer n will approach multiples of Pi, making sin(n) very close to 0. Though are there a's big enough for every value b to make the sum converge? How about other functions timed sin(n) power b? e.g(e^n*sin(n)^b)--128.237.206.250 (talk) 14:27, 22 July 2013 (UTC)
 * It should be the inverse of n^a*sin(n)^b. Sorry.--128.237.184.119 (talk) 17:04, 22 July 2013 (UTC)
 * An example Mathematica:ListPlot[FoldList[(#1 + 1./(#2^3 (Sin[#2]^2))) &, 0, Range[1000000]]]. It approaches 30 .3145 as the book says.--128.237.137.201 (talk) 04:04, 23 July 2013 (UTC)
 * If you consider the irrationality measure of π, μ, I think you can show that for almost all positive integers, $$\left|\frac{1}{\sin n}\right| < n^{\mu-1}$$, and so the sum should converge when $$a > 1 + b(\mu-1)$$. μ has a known upper bound, so in particular $$a = 16,\ b=2$$ should work. It probably also converges with some much lower values of a. -- Meni Rosenfeld (talk) 08:26, 23 July 2013 (UTC)