On the Hurwitz Zeta Function In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like
$\zeta(2k+1,\frac{1}{4})=C(k)\zeta(2k+1)$. With $C(k)$ some elementary function of $k$. 
After some basic calculations I found the following
$\zeta(2k+1,\frac{1}{4})=2^{2k+1}(2^{2k+1}-1)\zeta(2k+1)-\zeta(2k+1,\frac{3}{4})$  
but I don't know what to do about $\zeta(2k+1,\frac{3}{4})$. Trying the same method on this function, I stumble upon $\zeta(2k+1,\frac{1}{8})$. And I'm going in a loop getting nothing like my prof. said. Also I've looked through books and articles on the Hurwitz function and found nothing.     
 A: It can't be written as a scaled up version of the zeta function by elementary functions, however
$$\zeta(2k+1,\frac{1}{4})=2^{2k}(2^{2k+1}-1)\zeta(2k+1)+\frac{(-4)^kE_{2k}\pi^{2k+1}}{2(2k)!}$$
Where $E_k$ are euler numbers, with the first few
$E_0=1$
$E_2=-1$
$E_4=5$
$E_6=-61$
A: \begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{4}) \\
\zeta (2n+1,\frac{3}{4})%
\end{array}%
\right\} =2^{2n}(2^{2n+1}-1){\zeta }(2n+1) \\
\pm \frac{1}{2\pi }\left( 2n+2+4^{2n+2}\right) {\zeta }(2n+2)-2\sum
\limits_{l=0}^{n-1}4^{2n-2l}{{{\zeta }(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(1)}
\end{equation}
\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{3}) \\
\zeta (2n+1,\frac{2}{3})%
\end{array}%
\right\} =\frac{{3^{2n+1}-1}}{2}{\zeta (2n+1)} \\
\pm \frac{\sqrt{3}}{2\pi }\left( \left( 2n+2+3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}3^{2n-2l}{{{\zeta }(2n-2l)\zeta }}%
(2l+2)\right)%
\end{array}
\tag*{(2)}
\end{equation}
\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{6}) \\
\zeta (2n+1,\frac{5}{6})%
\end{array}%
\right\} =\frac{{6^{2n+1}-{3^{2n+1}}-{{2^{2n+1}}+1}}}{2}{\zeta (2n+1)} \\
\pm \frac{1}{2\sqrt{3}\pi }\left( 6^{2n+2}-3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}\left( 6^{2n-2l}-3^{2n-2l}\right) {{{\zeta }%
(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(3)}
\end{equation}
