Disprove convergence for $\sum_{n=1}^{\infty}{\frac{(-1)^{\lfloor nx \rfloor}}{n}}$

Question

A question found in a Russia exercise book, but with no answer.

Disrove $\sum_{n=1}^{\infty}{\frac{(-1)^{\lfloor nx \rfloor}}{n}}$ is uniform convergent in $(1,2)$.

Answer 1

Not a proof:

Since

$$\sum_{n=1}^{\infty}{\frac{(-1)^{\lfloor nx \rfloor}}{n}}=\left( \sum_{n=1}^{\infty}{(-1)^k} \right) \left( \sum_{k=\lfloor nx \rfloor}^{}{\frac{1}{n}} \right) =\left( \sum_{n=1}^{\infty}{(-1)^k} \right) \left( \sum_{k^2/x\leqslant n<(k+1)^2/x}^{}{\frac{1}{n}} \right),$$

we need to estimate $\sum_{k^2/x\leqslant n<(k+1)^2/x}^{}{\frac{1}{n}}$.

Use E-M Formula, we can get

$$\sum_{k^2/x\leqslant n<(k+1)^2/x}^{}{\frac{1}{n}}=\int_{k^2/x}^{(k+1)^2/x}{\frac{1}{n}}\mathrm{d}u-\left. \left( u-\lfloor u \rfloor -\frac{1}{2} \right) \frac{1}{u} \right|_{k^2/x}^{(k+1)^2/x}-\int_{k^2/x}^{(k+1)^2/x}{\left( u-\lfloor u \rfloor -\frac{1}{2} \right) \frac{1}{u^2}}\mathrm{d}u,$$

For the left side,

\begin{align*} |LHS|&\leqslant \ln \left( 1+\frac{1}{k} \right) +\left. \left( u-\lfloor u \rfloor -\frac{1}{2} \right) \frac{1}{u} \right|_{k^2/x}^{(k+1)^2/x}+\int_{k^2/x}^{(k+1)^2/x}{\left| u-\lfloor u \rfloor -\frac{1}{2} \right|\frac{1}{u^2}}\mathrm{d}u \\ &\leqslant \ln \left( 1+\frac{1}{k} \right) +x\left( \frac{1}{k^2}+\frac{1}{(k+1)^2} \right) +\int_{k^2/x}^{(k+1)^2/x}{\left| u-\lfloor u \rfloor -\frac{1}{2} \right|\frac{1}{u^2}}\mathrm{d}u \\ &\leqslant \ln \left( 1+\frac{1}{k} \right) +O(k^{-2})+O(k^{-3}). \end{align*}