For what real numbers α does (1/ 1 cosec(1) - 1)α + (1/2 cosec(1/2) - 1)α + ... + (1/n cosec(1/n) - 1)α + ... converge?
Solution
Answer: α > 1/2.
cosec 1/n is about n + 1/(6n). So the series is roughly 1/6 ∑1/n2α, which converges iff α > 1/2. We just have to tighten that up slightly.
n sin 1/n = 1 - 1/(6n2) + 1/(120n4) - ..., so 1/n cosec(1/n) = 1/(1 - 1/(6n2) + 1/(120n4) - ... ) = 1 + 1/(6n2) + terms in 1/n4 and higher. Hence (1/n cosec(1/n) - 1) = 1/n2 (1/6 + terms which tend to zero as n → ∞). So for n sufficiently large (1/n cosec(1/n) - 1) lies between 1/(12n2) and 1/(3n2) and hence the nth term of the series lies between k/n2α and k'/n2α. But ∑ 1/nβ converges for β > 1 and diverges for β ≤ 1. So the comparison test gives the result.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998