Does $\sum_{n=1}^\infty\frac{(-1)^{[\sqrt{nx}]}}{n}$ converges uniformly on $(0,1)$ or $(1,2)$?

Question

Does $\sum_{n=1}^\infty\frac{(-1)^{[\sqrt{nx}]}}{n}$ converges uniformly on $(0,1)$ or $(1,2)$?

$[\sqrt{nx}]$ stands for round down to integer. The most annoying part of this series is the $[\sqrt{nx}]$ part. I transformed it into $\frac{(-1)^n}{n}\cdot(-1)^{[\sqrt{nx}]-n}$ and tried A-D test, but I can't prove that the $(-1)^{[\sqrt{nx}]-n}$ part's sum is uniformly bounded. How could we deal with it? Any help would be appreciated.

Answer 1

We will prove that the series is uniformly convergent on any closed interval $I=[a,b]$ with $a>0$

Note that it is enough to prove that (with a constant depending only on $I$) there is $\delta >0$ st for all $N \ge 2$ say we have $$\sum_{N \le n \le N_1 }(-1)^{[\sqrt{nx}]}=O_I(N^{1-\delta}), x \in I, N \le N_1 \le 2N \tag 1$$

Then partial summation immediately implies that $$\sum_{N \le n \le N_1 }\frac{(-1)^{[\sqrt{nx}]}}{n}=O_I(N^{-\delta}), x \in I, N \le N_1 \le 2N$$

Hence if $N<M$ large, we use dyadic subdivision of the interval $[N,M]$ into $[N, 2N], [2N, 4N],...[2^kN, M]$ (where $2^kN < M \le 2^{k+1}N$) and get that $$|\sum_{N \le n \le M }\frac{(-1)^{[\sqrt{nx}]}}{n}| \le C_I N^{-\delta}(1+2^{-\delta}+...2^{-k\delta}) \le KC_IN^{-\delta}$$ since $\sum 2^{-m\delta}=K<\infty$ which implies the uniform convergence of the series on $I$ by the Cauchy criterion

To prove $(1)$, we will use the first derivative test for exponential sums and Vaaler trigonometrical polynomial approximation for the function $\psi(x)=x-[x]-1/2$ since $$(-1)^{[x]}=2\psi(x)-4\psi(x/2)$$ (which can be easily checked for $x \in [0,2)$ and then extended by periodicity as both sides have period $2$)

Vaaler approximation theorem (see Vaaler Some Extremal Functions in Fourier Analysis Page $210$) says that for every $J \ge 1$ there are coefficients $|a(j)| << 1/j$ st $$\psi(x) << 1/J+\sum_{1 \le |j| \le J}a(j)e(jx), e(x)=\exp {2\pi ix}$$ as usual and with all the implied constants universal.

This means that (for $N \le N_1 \le 2N$ as above) $$\sum_{N \le n \le N_1}\psi(\sqrt {nx}) << N/J+\sum_{1 \le j \le J}\frac{1}{j}|\sum_{N \le n \le N_1}e(j\sqrt {nx})| $$

Now the function $f(y)=(j\sqrt x) \sqrt y, y \in [N, 2N], x \in I, j \in [1, J]$ has derivative $\frac{j\sqrt x}{2\sqrt y}$ which is less than $1/2$ (and is $\approx \frac{j\sqrt x}{\sqrt N}$) for $J \le c_I N^{1/2}$ and so by the first derivative test for exponential sums (a simple statement and proof of it can be found on this useful blog) we have $$\sum_{N \le n \le N_1}e(j\sqrt {nx})|=O_I(\frac{\sqrt N}{j \sqrt x})$$ hence $$\sum_{1 \le j \le J}\frac{1}{j}|\sum_{N \le n \le N_1}e(j\sqrt {nx})| =O_I(\sqrt N)$$ since $1/\sqrt b \le 1/\sqrt x \le 1/\sqrt a, x \in I$ and valid as before for $J \le c_IN^{1/2}$

Now choosing $J=c_I N^{1/2}$ (so $N/J \approx \sqrt N$), it follows that: $$\sum_{N \le n \le N_1}\psi(\sqrt {nx})=O_I(N^{1/2})$$

Similarly $$\sum_{N \le n \le N_1}\psi(\sqrt {nx/2})=O_I(N^{1/2})$$ so putting it together we get $$|\sum_{N \le n \le N_1 }(-1)^{[\sqrt{nx}]}|\le 2|\sum \psi (\sqrt{nx})|+4|\sum \psi (\sqrt{nx/2})|=O_I(N^{1/2})$$ and the required bound $(1)$ is proved with $\delta=1/2$

Answer 2

$\begin{aligned} &\sum_{n\ge 1}\frac{(-1)^{[\sqrt{nx}]}}{n}\\ =&\sum_{k\ge1}(-1)^k\sum_{[\sqrt{nx}]=k}\frac 1n\\ =&\sum_{k\ge1}(-1)^k\sum_{\frac{k^2}{x}\le n<\frac{(k+1)^2}{x}}\frac 1n \end{aligned}$

Now let's estimate $\displaystyle\sum_{\frac{k^2}{x}\le n<\frac{(k+1)^2}{x}}\frac 1n$.

From Euler-Maclaurin's summation formula, we can get:

$\begin{aligned} &\sum_{\frac{k^2}{x}\le n<\frac{(k+1)^2}{x}}\frac 1n\\ =&\int_{\frac{k^2}{x}}^{\frac{(k+1)^2}{x}}\frac 1u \mathrm du-\frac{u-[u]-\frac12}{u}\bigg |^{\frac{(k+1)^2}{x}}_{\frac{k^2}{x}}-\int_{\frac{k^2}{x}}^{\frac{(k+1)^2}{x}}\frac{u-[u]-\frac12}{u^2}\mathrm du\\ &|\sum_{\frac{k^2}{x}\le n<\frac{(k+1)^2}{x}}\frac 1n|\\\ \le& \log(1+\frac1k)+x(\frac1{k^2}+\frac1{(k+1)^2})+\int_{\frac{k^2}{x}}^{\frac{(k+1)^2}{x}}|\frac{u-[u]-\frac12}{u^2}|\mathrm du\\ \le&\log(1+\frac1k)+O(\frac1{k^2})+O(\frac1{k^3}) \end{aligned}$

As Conrad pointed out in the comment, this estimation isn't accurate enough to prove the convergence. :(