For any x∈R, the series

∑n≥11nsin(xn)

is trivially absolutely convergent. It defines a function f(x) andI would like to show that f(x) is unbounded over R. Here there are my thoughts/attempts:

- (Lf)(s)=∑n≥111+n2s2=−s+πcothπs2s=∑m≥1(−1)m+1ζ(2m)s2m

is a function with no secrets. It behaves like π2s in a right neighbourhood of the origin, like π26s2 in a left neighbourhood of +∞. The origin is an essential singularity and there are simple poles at each s of the form ±im with m∈N+. These facts do not seem to rule out the possibility that f is bounded;- For any N∈N+ there clearly is some x≪eN such that sin(x),sin(x2),…,sin(xN) are all positive and large enough, making a partial sum of ∑n≥11nsin(xn) pretty close to ClogN. On the other hand I do not see an effective way for controlling ∑n>N1nsin(xn) – maybe by summation by parts, by exploiting the bounded-ness of the sine integral function?
- Some probabilistic argument might be effective. For any n≥3 we may define En as the set of x∈R+ such that sin(xn)≥1logn. The density of any En in R+ is close to 12, so by a Borel-Cantellish argument it looks reasonable that the set of points such that |f(x)|≥logx100 is unbounded, but how to make it really rigorous?
- To compute limx→x0f(x) through convolutions with approximate identities seems doable but not really appealing.

**Answer**

Thanks to MartinR and HJol for referencing a classical result of Hardy and Littlewood, which I am going to outline. The final result is

∑n≥11nsinxn=Ω(√loglogx) as x→+∞.

**Proof**:

Let q be a number of the form ∏pα, where each prime p is of the form 4n+1. Since \mathbb{Z}[i] is a UFD, the set of qs is the set of odd numbers which can be represented as a sum of two coprime squares. Let K=\prod_{q\leq 4k+1}q, x=\frac{\pi}{2}K and x_j=(4j+1)x for any j\in[1,K]. By introducing

Q^*(x) = \sum_{n=1}^{K}\frac{1}{n}\,\sin\frac{x}{n}

it is enough to approximate Q^* evaluated at each x_j. If n\mid K then \frac{K}{n} is a number of the form 4n+1 and \sin\left(\frac{x_j}{n}\right)=1, hence

Q^*(x_j) = \sum_{n\mid K}\frac{1}{n}+\sum_{\substack{1\leq n \leq K\\ n\nmid K}}\frac{1}{n}\,\sin\left(\frac{x_j}{n}\right)=\lambda(k)+R(x_j)

and

\frac{1}{K}\sum_{j=1}^{K}Q^*(x_j) = \lambda(k)+\frac{1}{K}\sum_{\substack{1\leq n \leq K\\ n\nmid K}}\frac{1}{n}\sum_{j=1}^{K}\sin\frac{(4j+1)x}{n}.

In the inner sum on the right, \frac{2x}{n} differs from \pi by at least \frac{C}{n}, such that \left|\sum_{j=1}^{K}\sin\frac{(4j+1)x}{n}\right|\ll\frac{1}{\left|\sin\frac{2x}{n}\right|}=O(n) and the LHS of the above equation is \geq \lambda(k)(1-\varepsilon). On the other hand it is trivial that

\lambda(k)\geq \sum_{q\leq 4k+1}\frac{1}{q}.

Let N(n) be the number of q which do not exceed n. By sieve methods it is well-known that N(n)\sim\frac{Dn}{\sqrt{\log n}}, hence the inequality

\sum_{q\leq 4k+1}\frac{1}{q}=\sum_{n\leq 4k+1}\frac{N(n)-N(n-1)}{n}\geq \sum_{n\leq 4k+1}\frac{N(n)}{n(n+1)}

finishes the proof.

Similarly it can be proved that

\sum_{n\geq 1}\frac{1}{n}\,\cos\frac{x}{n}=\Omega\left(\log\log x\right)\text{ as }x\to +\infty.\tag{CR}

**Alternative approaches**: to replace \sum_{n=m}^{M}\frac{1}{n}\sin\frac{x}{n} with a similarly shaped integral and to carefully estimate the error terms. A classical tool for performing such manipulations is the Denjoy-Koksma inequality. It can be combined with Van Der Corput’s trick and known inequalities in order to produce interesting generalizations, as shown by Flett and Codecà.

**Attribution***Source : Link , Question Author : Jack D’Aurizio , Answer Author : Jack D’Aurizio*