Proof of the unbounded-ness of ∑n≥11nsinxn\sum_{n\geq 1}\frac{1}{n}\sin\frac{x}{n}

For any xR, the series
is trivially absolutely convergent. It defines a function f(x) and I would like to show that f(x) is unbounded over R. Here there are my thoughts/attempts:

  1. (Lf)(s)=n111+n2s2=s+πcothπs2s=m1(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 mN+. These facts do not seem to rule out the possibility that f is bounded;
  2. For any NN+ there clearly is some xeN such that sin(x),sin(x2),,sin(xN) are all positive and large enough, making a partial sum of n11nsin(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?
  3. Some probabilistic argument might be effective. For any n3 we may define En as the set of xR+ 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?
  4. To compute limxx0f(x) through convolutions with approximate identities seems doable but not really appealing.


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

n11nsinxn=Ω(loglogx) as x+.


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)
\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.

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

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