Sums of form $\sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \cos(nx)$ Can we do the sums
\begin{align}
\sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \sin(nx)\\
\sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \cos(nx) 
\end{align}
They appear similar to known results e.g. this answer, this one and this one, but I can't solve it myself or find an answer.
From a numerical investigation, I find e.g.,
$$
\sum_{n=1}^{\infty} \frac{1}{n^{5/2}} \cos(nx) \approx - 3/4 \pi |\sin(x/2)| + \pi/2
$$
$$
\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{5/2}} \cos(nx) \approx - 3/4 \pi |\cos(x/2)| + \pi/2
$$
$$
\sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \sin(nx) \approx \pm \sin(x)
$$
Are these results exact? If not, what are the answers? and why are these approximations rather good?
 A: Only a hint:
Using the considerations in The sum of fractional powers $\sum\limits_{k=1}^x k^t$. I only found 
$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}(\sin(xk)+\cos(xk))=-\frac{(2\pi)^2}{3}\zeta(-\frac{3}{2},\frac{x}{2\pi})$$ 
and therefore
$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}\cos(xk)=-\frac{(2\pi)^2}{6}\left(\zeta(-\frac{3}{2},\frac{x}{2\pi})+\zeta(-\frac{3}{2},-\frac{x}{2\pi})\right)$$ 
and  
$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}\sin(xk)=-\frac{(2\pi)^2}{6}\left(\zeta(-\frac{3}{2},\frac{x}{2\pi})-\zeta(-\frac{3}{2},-\frac{x}{2\pi})\right)$$ 
so that one can compare $\enspace\displaystyle -\frac{(2\pi)^2}{6}\zeta(-\frac{3}{2},\frac{\pm x}{2\pi})\enspace$ with the assumptions for 
$\displaystyle \sum\limits_{k=1}^\infty\frac{\sin(xk)}{k^{5/2}}\enspace$ and $\enspace\displaystyle \sum\limits_{k=1}^\infty\frac{\cos(xk)}{k^{5/2}}$ .
To calculate the Hurwitz Zetafunction $\,\zeta(s,x)\,$ especially for negative $\,s\,$ see 
the different possibilities here: $\enspace$ https://en.wikipedia.org/wiki/Hurwitz_zeta_function 
