Convergence/Divergence of infinite series ∑∞n=1(sinn+2)nn3n\sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}

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 [1ε;1]) are somewhat 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 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 (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 |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 Cεk=Cε11μk

Since ε<1 and 11μ<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)[12ε;1ε]

Sε=(2+sin(n))nn3nk1(1ε/3)kC2εkC2ε=log(1(1ε/3)C2ε)C2εlog(1(1C2εε/3))C2ε=log(C2εε/3))C2ε

C2ε=C2ε11μ=Cεν with ν=12(1μ)]1/2;0[, so :

Sεlog(C/3)+(1+ν)logεCεν


Finally, we have to check if the series S2k converges or not :

S2klog(C/3)k(1+ν)log2C2kν=(A+Bk)(2ν)k

Since 2ν<1, the series converges.

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

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