Context of this problem: $\sum_{n\,\text{odd}} (-1)^{\frac{n-1}2}\frac{\log n}{\sqrt{n}} /\sum_{n\,\text{odd}}(-1)^{\frac{n-1}{2}}\frac{1}{\sqrt{n}}$ I remember seeing this somewhere a while ago - I'd given it a go but it was - and still is - beyond my capabilities. The problem came with the tag: "requires knowledge of analytic number theory". I am not necessarily asking for a solution. 
$$\left[\;\sum_{n\;\text{odd}} (-1)^{\frac{n-1}{2}}\frac{ \log n}{\sqrt{n}} \right]\left[\;\sum_{n \;\text{odd}} (-1)^{\frac{n-1}{2}}\frac{1}{\sqrt{n}} \right]^{-1}$$
Does anyone know where this is from? I am equally puzzled by the "hint". Does it have a number-theoretic interpretation that someone with basic (i.e. olympiad-level) knowledge of number theory could understand?
Would appreciate a reference if this appears in the literature somewhere.
 A: Hint. One may start with the analytic extension of the Hurwitz Riemann zeta function initially defined as
$$
\sum _{k=1}^{\infty } \frac{1}{(k+a)^s}=\zeta(s,a+1),\quad \Re s>1,\, \Re a>-1,\tag1
$$ giving the main result:

$$
\sum_{n\;\text{odd}} (-1)^{\frac{n-1}{2}}\frac1{n^s}=\sum _{k=1}^{\infty } \frac{(-1)^{k-1}}{(2k-1)^s}=2^{-2 s} \zeta\left(s,\frac14\right)-2^{-2 s} \zeta\left(s,\frac34\right). \tag2
$$

Thus

$$
\begin{align}
\sum_{n\;\text{odd}} (-1)^{\frac{n-1}{2}}\frac1{\sqrt{n}}=\sum _{k=1}^{\infty } \frac{(-1)^{k-1}}{(2k-1)^{1/2}}
=\frac12\zeta\left(\frac12,\frac14\right)-\frac12\zeta\left(\frac12,\frac34\right) 
\end{align}\tag3
$$ 

and, by differentiating $(2)$ with respect to $s$,

$$
\begin{align}
\sum_{n\;\text{odd}} (-1)^{\frac{n-1}{2}}\frac{ \log n}{\sqrt{n}}
=\ln 2\left(\zeta\left(\frac12,\frac14\right)-\zeta\left(\frac12,\frac34\right)\!\right)-\frac12\left(\zeta'\!\left(\frac12,\frac14\right)-\zeta'\!\left(\frac12,\frac34\!\right)\right) \tag4
\end{align}
$$ 

where $\displaystyle \zeta'\left(s_0,a\right)=\partial_s\left.\zeta\left(s,a\right)\right|_{s=s_0}.$
One may simplify it further using some special values of the Hurwitz zeta function.
