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

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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Just read from a book that it is not proven whether the case a=3 and b=2 converges or diverges(might be outdated, though), because sometimes the integer n will approach multiples of Pi, making sin(n) very close to 0.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)