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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.

tyobrien
  • 3,667

2 Answers2

4

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 $$ \sum\limits_{n = 1}^\infty { \frac{{( - 1)^n\log n}}{{2n + 1}}} = \int_0^{ + \infty } {\frac{1}{t}\left( {\left( {\frac{\pi }{4} - 1} \right)e^{ - t} - e^{\frac{t}{2}} \arctan ( e^{ - \frac{t}{2}} ) + 1} \right)dt} . $$

Gary
  • 36,640
1

I do not know if this could be of any help.

For large values of $x$, we have $$\zeta ^{(1,0)}(-1,x)=x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)-\frac{1}{2} x \log (x)+\frac{1}{12} (\log (x)+1)+$$ $$\frac{1}{720 x^2}-\frac{1}{5040 x^4}+\frac{1}{10080 x^6}-\frac{1}{9504 x^8}+\frac{691}{3603600 x^{10}}+O\left(\frac{1}{x^{12}}\right)$$ Using your fiirst formula, this would give $$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=n \log (2 n)+\frac{1}{4}\log (n)+\left(3 \log (A)-\frac{\log (2)}{12}\right)+\frac{1}{192 n^2}-$$ $$\frac{1}{1280 n^4}+\frac{17}{43008 n^6}-\frac{31}{73728 n^8}+\frac{691}{901120 n^{10}}+O\left(\frac{1}{n^{12}}\right)$$ which is almost exact even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 5 & 12.603993216766530412 & 12.603993216758928837 \\ 10 & 31.221522180442592703 & 31.221522180442590647 \\ 15 & 52.383497574013162289 & 52.383497574013162273 \\ 20 & 75.215036332678595144 & 75.215036332678595144 \end{array} \right)$$

Claude Leibovici
  • 289,558