∞∑n=1(sinn+2)nn3n

Does it converge or diverge?

Can we have a rigorous proof that is not probabilistic?

For reference, this question is supposedly a mix of real analysis and calculus.

**Answer**

The values for which sin(n) is close to 1 (say in an interval [1−ε;1]) are somewhat 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)|≤2√ε

If m>n and 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|≤4√ε 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,

|n−kπ|≥k1−μ.

By picking ε small enough we can forget about the finite number of exceptions to the inequality, and we get 4√ε≥(2k)1−μ.

Thus (m−n)≥2kπ−4√ε≥π(4√ε)11−μ−4√ε≥Aε=A√ε11−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 smaller (as we look for more problematic terms)

We can get a lower bound for the first problematic term using the irrationality measure as well : from |n−2k(n)π−π2|≤2√ε, we get that for ε small enough, (4k+1)1−μ≤|2n−(4k+1)π|≤4√ε, and then n≥Bε=B√ε11−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≤sinn is greater than Cεk=C√ε11−μk

Since ε<1 and 11−μ<0, this bound Cε grows when ε gets smaller.

And furthermore, the speed of this growth is greater if we can pick a smaller (better) value for μ (though all that matters is that μ is finite)

Now let us give an upper bound on the contribution of the terms where n is an integer such that sin(n)∈[1−2ε;1−ε]

Sε=∑(2+sin(n))nn3n≤∑k≥1(1−ε/3)kC2εkC2ε=−log(1−(1−ε/3)C2ε)C2ε≤−log(1−(1−C2εε/3))C2ε=−log(C2εε/3))C2ε

C2ε=C√2ε11−μ=C′εν with ν=12(1−μ)∈]−1/2;0[, so :

Sε≤−log(C′/3)+(1+ν)logεC′εν

Finally, we have to check if the series ∑S2−k converges or not :

∑S2−k≤∑−log(C′/3)−k(1+ν)log2C′2−kν=∑(A+Bk)(2ν)k

Since 2ν<1, the series converges.

**Attribution***Source : Link , Question Author : badreferences , Answer Author : mercio*