One can calculate a closed-form of (this or different examples) $$\sum_{k=1}^\infty\frac{C(2\sqrt{k})}{k^{5/2}},$$ where $C(x)$ is the Fresnel C integral from the Fourier expansion of the fractional part function $ \left\{ x \right\} $, multiplying by a factor and integrating: $$\int_0^1 (x-\frac{1}{2})\sqrt{x}dx=-\frac{1}{\pi}\sum_{k=1}^\infty\frac{1}{k}\int_0^1\sin(2\pi k x)\sqrt{x}dx.$$ From here you need this integration (also another in LHS) integrate sin(2pi kx) (sqrt(x)) dx, from x=0 to 1, this is the Wolfram Alpha code, and after do the sum. I don't know if this trick and series are well known or interesting (I am calculating and manipulating different examples), but I am interested in ask it here to answer this
Question. It's easy to prove with the definition of the Fresnel C integral that our series is convergent, and as I've said it is easy to get it in a closed-form, but imagine that we need answer the following exercise:
Do a study of the speed of convergence of our integral. That is, let $$\sum_{k=1}^\infty\frac{C(2\sqrt{k})}{k^{5/2}},$$ then study and convince us about if our series converges more or less quickly (you can do those comparisons that you need), and convince us if it's possible improve this speed of such convergence. That's is, how one does a rigurous study of the convergence of our series? Thanks in advance.
By the definition of $C(z)$ we have $$ C(2\sqrt{k}) = \int_{0}^{2\sqrt{k}}\cos\left(\tfrac{\pi x^2}{2}\right)\,dx = \int_{0}^{k}\frac{\cos(2\pi t)}{\sqrt{t}}\,dt = \sqrt{\frac{k}{2\pi}}\int_{0}^{2\pi}\frac{\cos(k u)}{\sqrt{u}}\,du$$ hence, in particular, $$ \sum_{k\geq 1}\frac{C(2\sqrt{k})}{k^{5/2}}=\frac{1}{\sqrt{2\pi}}\int_{0}^{2\pi}\left(\color{blue}{\sum_{k\geq 1}\frac{\cos(ku)}{k^2}}\right)\frac{du}{\sqrt{u}} $$ where (over the interval $(0,2\pi)$) the blue series is the Fourier series of a Bernoulli polynomial, $$ \sum_{k\geq 1}\frac{\cos(ku)}{k^2} = \frac{u^2}{4}-\frac{\pi u}{2}+\frac{\pi^2}{6} $$ leading, by uniform convergence, to: $$\sum_{k\geq 1}\frac{C(2\sqrt{k})}{k^{5/2}}= \frac{1}{\sqrt{2\pi}}\int_{0}^{2\pi}\left(\frac{u^{3/2}}{4}-\frac{\pi\sqrt{u}}{2}+\frac{\pi^2}{6\sqrt{u}}\right)\,du = \color{red}{\frac{\pi^2}{15}}.$$