TITLE: Closed form for limit of a sum $\lim\limits_{m\to\infty}(\ln m-4\sum_{n=1}^{m}\frac{(-1)^{n}}{2n+1}\sum_{k=1}^{2n}(-1)^{k}k\ln k)\approx-0.092$
QUESTION [5 upvotes]: I'm looking for a closed form of
$$\lim\limits_{m\to\infty}\left(\ln m-4\sum_{n=1}^{m}\frac{\left(-1\right)^{n}}{2n+1}\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k\right)\approx-0.092$$
for even $m$.
(For context this is a continuation of my ongoing work from this answer.)
Here's what I've tried so far. Wolfram alpha gives gives me
$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\zeta^{(1,0)}\left(-1,n+\frac{1}{2}\right)+2\zeta^{(1,0)}\left(-1,n+1\right)+n\ln2+3\ln A-\frac{1}{12}\ln2-\frac{1}{4}$$
From Wolfram's function site I get
$$\zeta^{(1,0)}(-1,x)=-\frac{1}{12}\left(-6x^{2}+18x-12\left(x-1\right)\ln\left(x-1\right)+12\ln A+6\left(x-1\right)\ln2\pi-13\right)+\psi^{(-2)}\left(x-1\right)$$
Combining these gets me
$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\psi^{(-2)}\left(n-\frac{1}{2}\right)+2\psi^{(-2)}\left(n\right)-\left(2n-1\right)\ln\left(n-\frac{1}{2}\right)+2n\ln n+n+n\ln2-\frac{1}{2}\ln2\pi+3\ln A-\frac{1}{12}\ln2-1$$
I could keep trying to find more expressions, but it feels like this is making things more complicated and possibly going to bring me back where I started.

REPLY [4 votes]: We have
$$
\zeta ( - 1,n + 1) =  - \frac{1}{2}n^2  - \frac{1}{2}n - \frac{1}{{12}}
$$
and
\begin{align*}
\zeta ^{(1,0)} ( - 1,n + 1) = \zeta ^{(1,0)} ( - 1,n) & + n\log n =  - \frac{1}{4}n^2  + \left( {\frac{1}{2}n^2  + \frac{1}{2}n + \frac{1}{{12}}} \right)\log n + \frac{1}{{12}} \\ & - \int_0^{ + \infty } {\left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} e^{ - nt} dt} 
\end{align*}
(cf. https://doi.org/10.1098/rspa.2017.0363). Noting that
$$
\zeta ^{(1,0)} \left( { - 1,n + \tfrac{1}{2}} \right) = \frac{{\log 2}}{2}\zeta ( - 1,2n + 1) + \frac{1}{2}\zeta ^{(1,0)} ( - 1,2n + 1) - \zeta ^{(1,0)} ( - 1,n + 1),
$$
we derive
\begin{align*}
 & \sum\limits_{k = 1}^{2n} {( - 1)^k k\log k}  = n\log (2n) + \frac{1}{4}\log n + 3\log A - \frac{{\log 2}}{{12}} \\ & + \int_0^{ + \infty } {\left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \left( {e^{ - 2nt}  - 4e^{ - nt} } \right)dt} .
\end{align*}
Thus, after some tedious calculations, we arrive at
\begin{align*}
& \mathop {\lim }\limits_{m \to  + \infty } \left( {\log (2m) - 4\sum\limits_{n = 1}^{2m} {\frac{{( - 1)^n }}{{2n + 1}}\sum\limits_{k = 1}^{2n} {( - 1)^k k\log k} } } \right)
\\ &
 = \mathop {\lim }\limits_{m \to  + \infty } \left[ {\log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n} } \right] \\ & + \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n \log n}}{{2n + 1}}}  + \left( {3\log A - \frac{{7\log 2}}{{12}}} \right)(4 - \pi )
\\ &
 + \int_0^{ + \infty } \!\! \left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \! \left( {16e^{\frac{t}{2}} \arctan (e^{ - \frac{t}{2}} ) - 4e^t \arctan (e^{ - t} ) - 12} \right)dt .
\end{align*}
Therefore, it remains to obtain a closed expression for
$$
\mathop {\lim }\limits_{m \to  + \infty } \left[ {\log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n} } \right].
$$
Some algebraic manipulation and Stirling's formula give
\begin{align*}
& \log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n}  =  - \log (2m) + 2\sum\limits_{n = 1}^{m - 1} {\log \left( {\frac{{2n + 1}}{{2n}}} \right)} \\ & =  - \log (2m) + 2\log \left( {\frac{{(2m - 1)!!}}{{(2m - 2)!!}}} \right) \to  \log \left( {\frac{2 }{\pi}} \right).
\end{align*}
Accordingly, a closed form expression for your limit is given by
\begin{align*}
&  \log \left( {\frac{2 }{\pi}} \right) + \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n \log n}}{{2n + 1}}}  + \left( {3\log A - \frac{{7\log 2}}{{12}}} \right)(4 - \pi )
\\ &
 + \int_0^{ + \infty } \!\! \left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \! \left( {16e^{\frac{t}{2}} \arctan (e^{ - \frac{t}{2}} ) - 4e^t \arctan (e^{ - t} ) - 12} \right)dt .
\end{align*}
The numerical value I got form this formula is $-0.09222744\ldots$. Using the formula
$$
\log n = \int_0^{ + \infty } {\frac{{e^{ - t}  - e^{ - nt} }}{t}dt} ,
$$
one can derive the quickly convergent integral representation
$$
\sum\limits_{n = 1}^\infty  { \frac{{( - 1)^n\log n}}{{2n + 1}}}  = \int_0^{ + \infty } {\frac{1}{t}\left( {\left( {\frac{\pi }{4} - 1} \right)e^{ - t}  - e^{\frac{t}{2}} \arctan  ( e^{ - \frac{t}{2}}  ) + 1} \right)dt} .
$$