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

Nyah Conrad

Nyah Conrad

Answered question


Does it converge or diverge?

Answer & Explanation



Beginner2022-02-24Added 8 answers

The values for which sin(n) is close to 1 (say in an interval [1ϵ; 1]) are some what regular:
1ϵsin(n) implies that there exists an integer k(n) such that n=2k(n)+π+π2+a(n) where |a(n)|arccos(1ϵ). As ϵ0, arccos(1ϵ)2ϵ, thus we can safely say that for ϵ small enough, |n2k(n)ππ2|=|a(n)|2ϵ
If m>n and since sin(n) and sin(m) are both in [1ϵ;1], then we have the inequality |(mn)2(k(m)k(n))π||m2k(m)ππ2|+|n2k(n)ππ2|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>2 such that for any integers n,k large enough, |nkπ|k1μ
By picking ϵ small enough we can forget about the finite number of exceptions to the inequality, and we get 4ϵ(2k)1μ. Thus
(mn)2kπ4ϵπ(4ϵ)11μ4ϵAϵ=Aϵ11μ 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 smaller
We can get a lower bound for the first problematic term using the irrationality measure as well : from |n2k(n)ππ2|2ϵ, we get that for ϵ small enough, (4k+1)1μ|2n(4k+1)π|4ϵ, and then nBϵ=Bϵ11μ for some constant B.
Therefore, there exists a constant C such that forall ϵ small enough, the k-th integer n such that 1ϵsinn is greater than Cϵ

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?