The discussion revolves around determining the convergence or divergence of the series \((-1)^n \frac{\cos^3(3^n x)}{3^n}\). Participants are exploring the applicability of the Leibniz Criterion and the implications of the cosine function's behavior in the series.
The discussion is ongoing, with various interpretations of the series being explored. Some participants have provided hints and suggestions for further investigation, while others express uncertainty about the implications of specific values of \(x\) on the series' behavior.
There is mention of specific values of \(x\) that may affect the nature of the series, including integer multiples of \(\pi\) and odd half-integer multiples of \(\pi\). Additionally, the series' behavior as \(n\) approaches infinity is under consideration, particularly regarding the terms that may converge to zero.
The series is not alternating, despite the factor of (-1)n, so I don't see that the Leibniz Criterion applies. I also don't see that the hint is helpful.peripatein said:Hi,
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
Is that really an x in the series?peripatein said:Hi,
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
Does the sum start at n=0 or n=1? That would account for the different results.peripatein said:I managed to express this series as a telescopic summation and thus merely (1/4)[-cos3x - ((1/3)^n)cos(3^(n+1)x)] remain in the sum.