Context of this problem: $\sum_{n\,\text{odd}} (-1)^{\frac{n-1}2}\frac{\log n}{\sqrt{n}} /\sum_{n\,\text{odd}}(-1)^{\frac{n-1}{2}}\frac{1}{\sqrt{n}}$

Question

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.

Answer 1

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.