Is $f(x) = \sum_{n\geq 1} \frac{\cos n x }{\sqrt{n}}$ monotonic on $(0,0.1)$?

Question

Is the function $f$ defined by$$f(x) = \sum_{n\geq 1} \frac{\cos n x }{\sqrt{n}}$$ monotonic on the interval $(0,0.1)$? By Dirichlet's test, the series converges on this interval.

Does it define a monotonically decreasing function?

I have tried to plot its graph. It seems that it is indeed monotonic. But as in numerics, I can only take a finite number of terms, the graph always displays some oscillation close to $x =0$ (the series is not uniformly converging on the interval), so I think decisive answer can come only from analytics.

This problem comes from my research.

I am curious whether some asymptotic analysis will be helpful.

Below is the graph of the function. I have taken 1000 terms.

Answer 1

Using a Riemann sum, $t=nx$ and $\mathrm{d}t=x$. As $x\to0$, $$ \begin{align} f(x) &=\sum_{n=1}^\infty\frac{\cos(nx)}{\sqrt{n}}\\ &=\frac1{\sqrt{x}}\sum_{n=1}^\infty\frac{\cos(nx)}{\sqrt{nx}}x\\ &\sim\frac1{\sqrt{x}}\int_0^\infty\frac{\cos(t)}{\sqrt{t}}\,\mathrm{d}t\\ &=\sqrt{\frac\pi{2x}} \end{align} $$ Using Riemann-Stieltjes integration: $$ \begin{align} f(x) &=\sum_{n=1}^\infty\frac{\cos(nx)}{\sqrt{n}}\\ &=\int_{0^+}^\infty\frac{\cos(tx)}{\sqrt{t}}\,\mathrm{d}(t-\{t\})\\ &=\int_{0^+}^\infty\frac{\cos(tx)}{\sqrt{t}}\,\mathrm{d}t -\int_{0^+}^\infty\frac{\cos(tx)}{\sqrt{t}}\,\mathrm{d}\{t\}\\ &=\frac1{\sqrt{x}}\int_0^\infty\frac{\cos(t)}{\sqrt{t}}\,\mathrm{d}t -\int_0^\infty\frac1{\sqrt{t}}\,\mathrm{d}\{t\}+O\!\left(x^2\right)\\ &=\sqrt{\frac\pi{2x}}+\zeta\!\left(\tfrac12\right)+O\!\left(x^2\right) \end{align} $$ If we continue in this fashion, we get $$ \sum_{n=1}^\infty\frac{\cos(nx)}{\sqrt{n}}=\sqrt{\frac\pi{2x}}+\zeta\!\left(\tfrac12\right)-\frac{\zeta\!\left(-\tfrac32\right)}2x^2+\frac{\zeta\!\left(-\tfrac72\right)}{24}x^4+O\!\left(x^6\right) $$

Answer 2

Using the integral representation of the polylogarithm, $$f(x) = \sum_{n \geq 1} \frac {\cos n x} {\sqrt n} = \frac 1 2 \operatorname{Li}_{1/2}(e^{i x}) + \frac 1 2 \operatorname{Li}_{1/2}(e^{-i x}) = \int_1^\infty \frac {t \cos x - 1} {t \sqrt {\pi \ln t \,} (1 - 2 t \cos x + t^2)} dt, \\ f(x_2) - f(x_1) = (\cos x_2 - \cos x_1) \int_1^\infty \frac {t^2 - 1} {\sqrt {\pi \ln t \,} (1 - 2 t \cos x_1 + t^2) (1 - 2 t \cos x_2 + t^2)} dt.$$ The last integrand is positive on $(1, \infty)$, therefore $f$ is monotonously decreasing on $(0, \pi)$.