∑n=1∞(sin⁡n+2)nn3nDoes it converge or diverge?

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poskakatiqzk · Accepted Answer

The values for which sin⁡(n) is close to 1 (say in an interval [1−ϵ; 1]) are some what regular:1−ϵ≤sin⁡(n) implies that there exists an integer k(n) such that n=2k(n)+π+π2+a(n) where |a(n)|≤arccos⁡(1−ϵ). As ϵ→0, arccos⁡(1−ϵ)∼2ϵ, thus we can safely say that for ϵ small enough, |n−2k(n)π−π2|=|a(n)|≤2ϵIf m&amp;gt;n and since sin⁡(n) and sin⁡(m) are both in [1−ϵ;1], then we have the inequality |(m−n)−2(k(m)−k(n))π|≤|m−2k(m)π−π2|+|n−2k(n)π−π2|≤4ϵ where (k(m)−k(n)) is some integer k.Since π has a finite irrationality measure, we know that there is a finite real constant μ&amp;gt;2 such that for any integers n,k large enough, |n−kπ|≥k1−μBy picking ϵ small enough we can forget about the finite number of exceptions to the inequality, and we get 4ϵ≥(2k)1−μ. Thus(m−n)≥2kπ−4ϵ≥π(4ϵ)11−μ−4ϵ≥Aϵ=Aϵ11−μ for some constant A.Therefore, we have a guarantee on the lengh of the gaps between equally problematic terms, and we know how this length grows as ϵ gets smallerWe can get a lower bound for the first problematic term using the irrationality measure as well : from |n−2k(n)π−π2|≤2ϵ, we get that for ϵ small enough, (4k+1)1−μ≤|2n−(4k+1)π|≤4ϵ, and then n≥Bϵ=Bϵ11−μ for some constant B.Therefore, there exists a constant C such that forall ϵ small enough, the k-th integer n such that 1−ϵ≤sin⁡n is greater than Cϵ

\sum_{n=1}^\infty\frac{(\sin n+2)^n}{n3^n} Does it converge or diverge?

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