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

 A: 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
$$
\sum_{n=1}^\infty\frac{\cos(nx)}{\sqrt{n}}=\sqrt{\frac\pi{2x}}+\zeta\!\left(\tfrac12\right)-\frac{\zeta\!\left(-\tfrac32\right)}2x^2+\frac{\zeta\!\left(-\tfrac72\right)}{24}x^4+O\!\left(x^6\right)
$$
A: 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 $(0, \pi)$.
